Question 15 Marks
Represent the following families of curves by forming the corresponding differential equation:
$\frac{\text{x}^2}{\text{a}^2}-\frac{\text{y}^2}{\text{b}^2}=1$
AnswerThe equation of the family of curves is
$\frac{\text{x}^2}{\text{a}^2}-\frac{\text{y}^2}{\text{b}^2}=1$
where a is a parameter.
As this equation has only one arbitrary constant, we shall get a differential equation of first order.
Differentiating (1) with respect to x, we get
$\frac{2\text{x}}{\text{a}^2}-\frac{2\text{y}}{\text{b}^2}\frac{\text{dy}}{\text{dx}}=0\ ...(2)$
Differentiating (2) with respect to x, we get
$\frac{2}{\text{a}^2}-\frac{2}{\text{b}^2}\Big(\frac{\text{dy}}{\text{dx}}\Big)^2-\frac{2\text{y}}{\text{b}^2}\frac{\text{d}^2\text{y}}{\text{dx}^2}=0$
$\Rightarrow\frac{2}{\text{a}^2}=\frac{2}{\text{b}^2}\Big[\text{y}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2\Big]$
$\Rightarrow\frac{\text{b}^2}{\text{a}^2}=\Big[\text{y}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2\Big]\ ...(3)$
Now, from (2), we get
$\frac{2\text{x}}{\text{a}^2}=\frac{2\text{y}}{\text{b}^2}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\frac{\text{b}^2}{\text{a}^2}=\frac{\text{y}^2}{\text{x}}\frac{\text{dy}}{\text{dx}}$
From (3), (4), we get
$\frac{\text{y}}{\text{x}}\frac{\text{dy}}{\text{dx}}=\Big[\text{y}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2\Big]$
$\Rightarrow\text{x}\Big[\text{y}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2\Big]=\text{y}\frac{\text{dy}}{\text{dx}}$
It is the required differential equation.
View full question & answer→Question 25 Marks
Solve the following differential equation:
$\frac{\text{dy}}{\text{dx}}+2\text{y}=\sin\text{x}$
AnswerWe have,
$\frac{\text{dy}}{\text{dx}}+2\text{y}=\sin\text{x}\ \dots(1)$
Clearly, it is a linear differential equation of the form
$\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
where
$\text{P}=1$
and
$\text{Q}=\sin\text{x}$
$\therefore$ I.F. $=\text{e}^{\int\text{Pdx}}$
$=\text{e}^{\int2\text{dx}}$
$=\text{e}^{2\text{x}}$
Multiplying both sides of (1) by $I.F. = e^{2x}$, we get
$\text{e}^{2\text{x}}\Big(\frac{\text{dy}}{\text{dx}}+2\text{y}\Big)=\text{e}^{2\text{x}}\sin\text{x}$
$\Rightarrow\ \text{e}^{2\text{x}}\frac{\text{dy}}{\text{dx}}+2\text{e}^{2\text{x}}\text{y}=\text{e}^{2\text{x}}\sin\text{x}$
Integrating both sides with respect to x, we get
$\text{ye}^{2\text{x}}=\int\text{e}^{2\text{x}}\sin\text{x dx + C}$
$\Rightarrow\ \text{ye}^{2\text{x}}=\frac{1}5\int[2\text{e}^{2\text{x}}(2\sin\text{x}-\cos\text{x}+\text{e}^{2\text{x}}(2\cos\text{x}+\sin\text{x})]$
Putting $\text{e}^{2\text{x}}(2\sin\text{x}-\cos\text{x})=\text{t}$
$\Rightarrow[2\text{e}^{2\text{x}}(2\sin\text{x}-\cos\text{x})+\text{e}^{2\text{x}}(2\cos\text{x}+\sin\text{x})]\text{dx}=\text{dt}$
$\therefore\ \text{ye}^{2\text{x}}=\frac{1}5\int\text{dt + C}$
$\Rightarrow\text{ye}^{2\text{x}}=\frac{\text{t}}5+\text{C}$
$\text{ye}^{2\text{x}}=\frac{\text{e}^{2\text{x}}}5(2\sin\text{x}-\cos\text{x})+\text{C}$
$\Rightarrow\ \text{y}=\frac{1}5(2\sin\text{x}-\cos\text{x})+\text{Ce}^{-2\text{x}}$
Hence, $\text{y}=\frac{1}5(2\sin\text{x}-\cos\text{x})+\text{Ce}^{-2\text{x}}$ is the required solution.
View full question & answer→Question 35 Marks
Form the differential equation of the family of circle in the secound qudrant and touching the coordinate axes.
AnswerLet C denote the family of circles in the second qwdrant and touching the coordinate axes.
Let (-a, a) be the coordinate of the centre of any member of this family,
Equation representing the family C is
$(\text{x}+\text{a})^2+(\text{y}-\text{a})^2=\text{a}^2\ ...(1)$
$\text{x}^2+\text{y}^2+2\text{ax}-2\text{ay}+\text{a}^2=0\ ...(2)$
Differentiating eqn (i) w.r.t.x, we get
$2\text{x}+2\text{y}\frac{\text{dy}}{\text{dx}}+2\text{a}-2\text{a}\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\text{x}+\text{y}\frac{\text{dy}}{\text{dx}}=\text{a}\Big(\frac{\text{dy}}{\text{dx}}-1\Big)$
$\Rightarrow\text{a}=\frac{\text{x}+\text{yy'}}{\text{y}-1}$
Substituting the value of a in (ii), we get
$\Big[\text{x}+\frac{\text{x}+\text{yy}'}{\text{y}-1}\Big]^2+\Big[\text{y}-\frac{\text{x}+\text{yy}'}{\text{y}-1}\Big]^2=\Big[\frac{\text{x}+\text{yy}'}{\text{y}-1}\Big]^2$
$\Rightarrow[\text{xy}-\text{x}+\text{x}+\text{yy'}]^2+[\text{yy}'-\text{y}-\text{x}-\text{yy}']^2=[\text{x}+\text{yy'}]^2$
$\Rightarrow(\text{x}+\text{y})^2\text{y}^2+(\text{x}+\text{y})^2=[\text{x}+\text{yy'}]$
$(\text{x}+\text{y})^2\Big[(\text{y})^2+1\Big]=[\text{x}+\text{yy'}]^2$
which is the differential equation representing the given family of circles.
View full question & answer→Question 45 Marks
Solve the following initial value problems:
$(1+\text{y}^2)\text{dx}+(\text{x}-\text{e}^{\tan^{-1}\text{y}})\text{dy}=0,\text{ y}(0)=0$
AnswerWe have,
$(1+\text{y}^2)\text{dx}+(\text{x}-\text{e}^{\tan^{-1}\text{y}})\text{dy}=0$
$\Rightarrow(\text{x}-\text{e}^{\tan^{-1}\text{y}})\frac{\text{dy}}{\text{dx}}=-(1+\text{y}^2)$
$\Rightarrow(1+\text{y}^2)\frac{\text{dx}}{\text{dy}}=-(\text{x}-\text{e}^{\tan^{-1}\text{y}})$
$\Rightarrow\frac{\text{dx}}{\text{dy}}+\frac{\text{x}}{1+\text{y}^2}=\frac{\text{e}^{-\tan^{-1}\text{y}}}{1+\text{y}^2}\ ...(1)$
Clearly, it is a linear differential equation of the form
$\frac{\text{dy}}{\text{dx}}+\text{Px}=\text{Q}$
Where $\text{P}=\frac{1}{1+\text{y}^2}$ and $\text{Q}=\frac{\text{e}^{-\tan^{-1}\text{y}}}{1+\text{y}^2}$
$\therefore\text{ I.F.}=\text{e}^{\int\text{Pdy}}$
$=\text{e}^{\int\frac{1}{1+\text{y}^2}\text{dy}}$
$=\text{e}^{\tan^{-1}\text{y}}$
Multiplying both sides of (1) by $\text{I.F.}=\text{e}^{\tan^{-1}\text{y}},$ we get
$\text{e}^{\tan^{-1}\text{y}}\Big(\frac{\text{dx}}{\text{dy}}+\frac{\text{x}}{1+\text{y}^2}\Big)=\text{e}^{\tan^{-1}\text{y}}\frac{\text{e}^{-\tan^{-1}\text{y}}}{1+\text{y}^2}$
$\Rightarrow\text{e}^{\tan^{-1}\text{y}}\Big(\frac{\text{dx}}{\text{dy}}+\frac{\text{x}}{1+\text{y}^2}\Big)=\frac{1}{1+\text{y}^2}$
Integrating both sides with respect to y, we get
$\text{e}^{\tan^{-1}\text{y}}\text{x}=\int\frac{1}{1+\text{y}^2}\text{dy}+\text{C}$
$\Rightarrow\text{x}\text{e}^{\tan^{-1}\text{y}}=\tan^{-1}\text{y}+\text{C}\ ...(2)$
Now,
$\text{y}(0)=0$
$\therefore\ 0\times\text{e}^{0}=0+\text{C}$
$\Rightarrow\text{C}=0$
Putting the value of C in (2), we get
$\text{x}\text{e}^{\tan^{-1}\text{y}}=\tan^{-1}\text{y}+0$
$\Rightarrow\text{x}\text{e}^{\tan^{-1}\text{y}}=\tan^{-1}\text{y}$
Hence, $\text{x}\text{e}^{\tan^{-1}\text{y}}=\tan^{-1}\text{y}$ is the required solution.
View full question & answer→Question 55 Marks
Form the differential equation of the family of curves represented by the equation (a being the perimeter):$(2\text{x}-\text{a})^2-\text{y}^2=\text{a}^2$
AnswerThe equation of the family of curves is
$(2\text{x}-\text{a})^2-\text{y}^2=\text{a}^2 ...(1)$
where a is a parameter.
As this equation has only one parameter, we shall get a differential equation of first order.
Differentiating (1) with respect to x, we get
$8\text{x}-4\text{a}-2\text{y}\frac{\text{dy}}{\text{dx}}=0\ ...(2)$
$\Rightarrow-\text{y}\frac{\text{dy}}{\text{dx}}+4\text{x}=2\text{a}$
Now, from (1), we get
$2\text{a}=\frac{4\text{x}^2-\text{y}^2}{2\text{x}}\ ...(3)$
From (2) and (3), we get
$-\text{y}\frac{\text{dy}}{\text{dx}}+4\text{x}=\frac{4\text{x}^2-\text{y}^2}{2\text{x}}$
$\Rightarrow-2\text{xy}\frac{\text{dy}}{\text{dx}}+8\text{x}^2=4\text{x}^2-\text{y}^2$
$\Rightarrow-2\text{xy}\frac{\text{dy}}{\text{dx}}+4\text{x}^2+\text{y}^2=0$
$\Rightarrow2\frac{\text{dy}}{\text{dx}}=4\text{x}^2+\text{y}^2$
It is the required differential equation.
View full question & answer→Question 65 Marks
A curve is such that the length of the perpendicular from the origin on the tangent at any point P of the curve is equal to the abscissa of P. Prove that the differential equation of the curve is $\text{y}^{2}-2\text{xy}\frac{\text{dy}}{\text{dx}}-\text{x}^{2}=0$ and hence find the curve.
Answer$\text{y}^{2}-2\text{xy}\frac{\text{dy}}{\text{dx}}-\text{x}^{2}=0$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{y}^{2}-\text{x}^{2}}{2\text{xy}}$
It is a homeganeous equation,
Put, $\text{y}=\text{vx}$
$\frac{\text{dy}}{\text{dx}}=\text{v}+\text{x}\frac{\text{dv}}{\text{dx}}$
Now,
$\text{x}\frac{\text{dv}}{\text{dx}}+\text{v}=\frac{\text{v}^{2}\text{x}^{2}-\text{x}^{2}}{2\text{xvx}}$
$\text{x}\frac{\text{dv}}{\text{dx}}=\frac{\text{v}^{2}-\text{1}}{2\text{v}}-\text{v}$
$\text{x}\frac{\text{dv}}{\text{dx}}=\frac{\text{v}^{2}-\text{1}-2\text{v}^{2}}{2\text{v}}$
$\text{x}\frac{\text{dv}}{\text{dx}}=\frac{-\text{v}^{2}-\text{1}}{2\text{v}}$
$\int\frac{2\text{v}}{\text{v}^{2}+1}\text{dv}=-\int\frac{\text{dx}}{\text{x}}$
$\log|\text{v}^{2}+1|=\log|\text{x}|+\log|\text{C}|$
$\text{v}^{2}+1=\frac{\text{C}}{\text{x}}$
$\frac{\text{y}^{2}+\text{x}^{2}}{\text{x}^{2}}=\frac{\text{C}}{\text{x}}$
$\text{y}^{2}+\text{x}^{2}=\text{Cx}$
$\text{y}^{2}+\text{x}^{2}-\text{Cx}=0$
Differntiating it with respect to X,
$2\text{x}+2\text{y}\frac{\text{dy}}{\text{dx}}-\text{C}=0$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{C}-2\text{x}}{2\text{y}}$
Let (h, k) be the point where tangent passes through origin and length is equal to h, So, equation of tangent at (h, k) is,
$(\text{y}-\text{k})=\Big(\frac{\text{dy}}{\text{dx}}\Big)_{(\text{h},\text{k})}(\text{x}-\text{h})$
$(\text{y}-\text{k})=\Big(\frac{\text{C}-2\text{h}}{2\text{k}}\Big)(\text{x}-\text{h})$
$2\text{ky}-2\text{k}^{2}=\text{xC}-2\text{hx}-\text{hC}+2\text{h}^{2}$
$\text{x}(\text{C}-2\text{h})-2\text{ky}+2\text{k}^{2}-\text{hc}+2\text{h}^{2}=0$
$\text{x}(\text{C}-2\text{h})-2\text{ky}+2(\text{k}^{2}+\text{h}^{2})-\text{hc}=0$
$\text{x}(\text{C}-2\text{h})-2\text{ky}+2(\text{Ch})-\text{hC}=0$
$\text{x}(\text{C}-2\text{h})-2\text{ky}+\text{hC}=0$
Lenght of perpendicular as tangent feom origin is
$\text{L}=\Big|\frac{\text{ax}_{1}+\text{by}_{1}+\text{C}}{\sqrt{\text{a}^{2}+\text{b}^{2}}}\Big|$
$\text{L}=\Big|\frac{(0)(\text{C}-2\text{h})+(0)(-2\text{k})+\text{hc}}{\sqrt{(\text{C}-2\text{h})^{2}+(-2\text{k})^{2}}}\Big|$
$\text{L}=\Big|\frac{\text{hc}}{\sqrt{\text{C}^{2}+4\text{h}^{2}+4\text{k}^{2}-4\text{Ch}}}\Big|$
$\text{L}=\frac{\text{hc}}{\sqrt{\text{C}^{2}+4(0)}}$
$\text{L}=\frac{\text{hC}}{\text{C}}$
Hence,
$\text{x}^{2}+\text{y}^{2}=\text{Cx}$ is the required curve.
View full question & answer→Question 75 Marks
Solve the following differential equation:
$(\text{x}^2-1)\frac{\text{dy}}{\text{dx}}+2(\text{x}+2)\text{y}=2(\text{x}+1)$
AnswerWe have,
$(\text{x}^2-1)\frac{\text{dy}}{\text{dx}}+2(\text{x}+2)\text{y}=2(\text{x}+1)$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}+\frac{2(\text{x}+2)}{\text{x}^2-1}\text{y}=\frac{2(\text{x}+1)}{\text{x}^2-1}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}+\frac{2(\text{x}+2)}{\text{x}^2-1}\text{y}=\frac{2}{\text{x}-1}\ \dots(1)$
Clearly, it is a linear differential equation of the form
$\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
where
$\text{P}=\frac{2(\text{x}+2)}{\text{x}^2-1}$
$\text{Q}=\frac{2}{\text{x}-1}$
$\therefore$ I.F. $=\text{e}^{\int\text{Pdx}}$
$=\text{e}^{\int\frac{2(\text{x+2})}{\text{x}^2-1}\text{dx}}$
$=\text{e}^{\int\frac{2\text{x}}{\text{x}^2-1}+4\int\frac{1}{\text{x}^2-1}\text{dx}}$
$=\text{e}^{\log|\text{x}^2-1|+4\times\frac{1}2\log\big|\frac{\text{x}-1}{\text{x}+1}\big|}$
$=\text{e}^{\log\Big|(\text{x}^2-1)\times\frac{(\text{x}-1)^2}{(\text{x}+1)^2}\Big|}$
$=\text{e}^{\log\Big|(\text{x}^2-1)\times\frac{(\text{x}-1)^2}{(\text{x}+1)^2}\Big|}$
$=\text{e}^{\log\Big|\frac{(\text{x}-1)^3}{(\text{x}+1)}\Big|}$
$=\frac{(\text{x}-1)^3}{(\text{x}+1)}$
Multiplying both sides of (1) by $\frac{(\text{x}-1)^3}{(\text{x}+1)},$ we get
$\frac{(\text{x}-1)^3}{(\text{x}+1)}\Big(\frac{\text{dy}}{\text{dx}}+\frac{2(\text{x}+2)}{\text{x}^2-1}\text{y}\Big)=\frac{(\text{x}-1)^3}{(\text{x}+1)}\times\frac{2}{\text{x}-1}$
$\Rightarrow\ \frac{(\text{x}-1)^3}{(\text{x}+1)}\frac{\text{dy}}{\text{dx}}-\frac{2(\text{x}+2)(\text{x}-1)^2}{(\text{x}+1)^2}\text{y}=\frac{2(\text{x}-1)^2}{(\text{x}+1)}$
Integrating both sides with respect to x, we get
$\frac{(\text{x}-1)^3}{(\text{x}+1)}\text{y}=\int\frac{2(\text{x}-1)^2}{(\text{x}+1)}\text{dx + C}$
$\Rightarrow\ \frac{(\text{x}-1)^3}{(\text{x}+1)}\text{y}=\int2\Big\{\frac{(\text{x}-1)^2-4\text{x}}{(\text{x}+1)}\Big\}\text{dx + C}$
$\Rightarrow\ \frac{(\text{x}-1)^3}{(\text{x}+1)}\text{y}=\int2\Big\{(\text{x}+1)-\frac{4\text{x}}{(\text{x}+1)}\Big\}\text{dx + C}$
$\Rightarrow\ \frac{(\text{x}-1)^3}{(\text{x}+1)}\text{y}=\int2\Big\{(\text{x}+1)-\frac{4(\text{x}+1-1)}{(\text{x}+1)}\Big\}\text{dx + C}$
$\Rightarrow\ \frac{(\text{x}-1)^3}{(\text{x}+1)}\text{y}=\int2\Big\{(\text{x}+1)-4+\frac{4}{(\text{x}+1)}\Big\}\text{dx + C}$
$\Rightarrow\ \frac{(\text{x}-1)^3}{(\text{x}+1)}\text{y}=\int\Big(2\text{x}-6+\frac{8}{\text{x}+1}\Big)\text{dx + C}$
$\Rightarrow\ \frac{(\text{x}-1)^3}{(\text{x}+1)}\text{y}=\text{x}^2-6\text{x}+8\log|\text{x}+1|+\text{C}$
$\Rightarrow\ \text{y}=\frac{(\text{x}-1)^3}{(\text{x}+1)}(\text{x}^2-6\text{x}+8\log|\text{x}+1|+\text{C})$
Hence, $\text{y}=\frac{(\text{x}-1)^3}{(\text{x}+1)}(\text{x}^2-6\text{x}+8\log|\text{x}+1|+\text{C})$ is the required solution.
View full question & answer→Question 85 Marks
Solve the following initial value problems $\tan\text{x}\Big(\frac{\text{dy}}{\text{dx}}\Big)=2\text{x}\tan\text{x}+\text{x}^2-\text{y},\tan\text{x}\neq0$ given that y = 0 when $\text{x}=\frac{\pi}{2}$
AnswerWe have,
$\tan\text{x}\Big(\frac{\text{dy}}{\text{dx}}\Big)=2\text{x}\tan\text{x}+\text{x}^2-\text{y}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}+\frac{1}{\tan\text{x}}\text{y}=\frac{2\text{x}\tan\text{x}+\text{x}^2}{\tan\text{x}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}+(\cot\text{x})\text{y}=2\text{x}+\text{x}^2\cot\text{x}$
This is a linear differential equation of the form $\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
Integrating factor,
$\text{ I.F.}=\text{e}^{\int\text{Pdx}}$
$=\text{e}^{\int\cot\text{x dx}}$
$=\text{e}^{\log\sin\text{x}}$
$=\sin\text{x}$
The solution of the given differential equation is given by
$\text{y}\times(\text{I.F.})=\int\text{Q}\times(\text{I.F.})\text{ dx}+\text{C}$
$\Rightarrow\text{y}\times\sin\text{x}=\int(2\text{x}+\text{x}^2\cot\text{x})\sin\text{x dx}+\text{C}$
$\Rightarrow\text{y}\sin\text{x}=\int2\text{x}\sin\text{x dx}+\int\text{x}^2\cos\text{x dx}+\text{C}$
$\Rightarrow\text{y}\sin\text{x}=\int2\text{x}\sin\text{x dx}+\Big[\text{x}^2\int\cos\text{x dx}-\int\Big(\frac{\text{d}}{\text{dx}}\text{x}^2\times\int\cos\text{x dx} \Big)\text{dx}\Big]$
$\Rightarrow\text{y}\sin\text{x}=\int2\text{x}\sin\text{x dx} +\text{x}^2\sin\text{x}-\int2\text{x}\sin\text{x dx}+\text{C}$
$\Rightarrow\text{y}\sin\text{x}=\text{x}^2\sin\text{x}+\text{C}$
$\Rightarrow\text{y}=\text{x}^2+\text{coses x}\times\text{C}\ ....(1)$
It is given that y = 0 when $\text{x}=\frac{\pi}{2}$
$\therefore\ 0=\big(\frac{\pi}{2}\Big)^2+\text{coses}\frac{\pi}{2}\times\text{C}$
$\Rightarrow\text{C}=-\frac{\pi^2}{4}$
Puttuing $\text{C}=-\frac{\pi^2}{4}$ in (1) we get
$\text{y}=\text{x}^2-\frac{\pi^2}{4}\text{coses x}$
Hence, $\text{y}=\text{x}^2-\frac{\pi^2}{4}\text{coses x}$ is the required solution.
View full question & answer→Question 95 Marks
Solve the following initial value problems:
$\frac{\text{dy}}{\text{dx}}+\text{y}\tan\text{x}=2\text{x}+\text{x}^2\tan\text{x},\text{ y}(0)=1$
AnswerWe have,
$\frac{\text{dy}}{\text{dx}}+\text{y}\tan\text{x}=2\text{x}+\text{x}^2\tan\text{x}\ ...(1)$
Clearly, it is a linear differential equation of the form
$\frac{\text{dy}}{\text{dx}}+\text{Px}=\text{Q}$
Where $\text{P}=\tan\text{x}$ and $\text{Q}=\text{x}^2\cot\text{x}+2\text{x}$
$\therefore\text{ I.F.}=\text{e}^{\int\text{Pdx}}$
$=\text{e}^{\int\tan\text{x dx}}$
$=\text{e}^{\log|\sec\text{x}|}$
$=\sec\text{x}$
Multiplying both sides of (1) by $\text{I.F.}=\sec\text{x},$ we get
$\sec\text{x}\Big(\frac{\text{dy}}{\text{dx}}+\text{y}\tan\text{x}\Big)=\sec\text{x}(\text{x}^2\tan\text{x}+2\text{x})$
$\Rightarrow\sec\text{x}\Big(\frac{\text{dy}}{\text{dx}}+\text{y}\tan\text{x}\Big)=\text{x}^2\tan\text{x }\sec\text{x}+2\text{x}\sec\text{x}$
Integrating both sides with respect to x, we get
$\text{y}\sec\text{x}=\int\text{x}^2\tan\text{x }\sec\text{x dx}+2\int\text{x}\sec\text{x dx}\\+2\int\text{x}\sec\text{x dx} +\text{C}$
$\Rightarrow\text{y}\sec\text{x}=\int\text{x}^2\tan\text{x }\sec\text{x dx}+2\sec\text{x}\int\text{x dx}\\-2\int\Big[\frac{\text{d}}{\text{dx}}(\sec\text{x})\int\text{x dx}\Big]\text{dx}+\text{C}$
$\Rightarrow\text{y}\sec\text{x}=\int\text{x}^2\tan\text{x }\sec\text{x dx}+\text{x}^2\sec\text{x}\\-\int\text{x}^2\tan\text{x }\sec\text{x dx} +\text{C}$
$\Rightarrow\text{y}\sec\text{x}=\text{x}^2\sec\text{x}+\text{C}$
$\Rightarrow\text{y}=\text{x}^2+\text{C}\cos\text{x}\ ...(2)$
Now,
$\text{y}(0)=1$
$\therefore\ 1=0+\text{C}\cos0$
$\Rightarrow\text{C}=1$
Putting the value of C in (2), we get
$\text{y}=\text{x}^2+\cos\text{x}$
Hence, $\text{y}=\text{x}^2+\cos\text{x}$ is the required solution.
View full question & answer→Question 105 Marks
Solve the following differential equation
$(\text{x}^3+\text{x}^2+\text{x}+1)\frac{\text{dy}}{\text{dx}}=2\text{x}^2+\text{x}$
AnswerWe have,
$(\text{x}^3+\text{x}^2+\text{x}+1)\frac{\text{dy}}{\text{dx}}=2\text{x}^2+\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{2\text{x}^2+\text{x}}{\text{x}^3+\text{x}^2+\text{x}+1}$
$\Rightarrow\text{dy}=\frac{2\text{x}^2+\text{x}}{(\text{x}+1)(\text{x}^2+1)}\ \text{dx}$
Intregrating both sides, we get
$\int\text{dy}=\int\Big\{\frac{2\text{x}^2+\text{x}}{(\text{x}+1)(\text{x}^2+1)}\Big\}\text{dx}$
$\Rightarrow\text{y}=\int\Big\{\frac{2\text{x}^2+\text{x}}{(\text{x}+1)(\text{x}^2+1)}\Big\}\text{dx}$
Let $\frac{2\text{x}^2+\text{x}}{(\text{x}+1)(\text{x}^2+1)}=\frac{\text{A}}{\text{x}+1}+\frac{\text{Bx}+\text{C}}{\text{x}^2+1}$
$\Rightarrow 2 x^2+x=A x^2+A+B x^2+B x+C x+C $
$\Rightarrow 2 x^2+x=(A+B) x^2+(B+C) x+(A+C)$
Compairing the coefficient on both sides, we get
$A + B = 2 ...(1)$
$B + C = 1 ...(2)$
$A + C = 0 ...(3)$
Solving (1), (2) and (3), we get
$\text{A}=\frac{1}{2}$
$\text{B}=\frac{3}{2}$
$\text{C}=-\frac{1}{2}$
$\therefore\text{y}=\frac{1}{2}\int\frac{1}{(\text{x}+1)}\ \text{dx}+\int\frac{\frac{3}{2}\text{x}-\frac{1}{2}}{\text{x}^2+1}\ \text{dx}$
$=\frac{1}{2}\int\frac{1}{(\text{x}+1)}\text{dx}+\frac{3}{4}\int\frac{2\text{x}}{\text{x}^2+1}\ \text{dx}-\frac{1}{2}\int\frac{1}{\text{x}^2+1}\ \text{dx}$
$=\frac{1}{2}\log|\text{x}+1|+\frac{3}{4}\log|\text{x}^2+1|-\frac{1}{2}\tan^{-1}\text{x}+\text{C}$
Hence, $\text{y}=\frac{1}{2}\log|\text{x}+1|+\frac{3}{4}\log|\text{x}^2+1|-\frac{1}{2}\tan^{-1}\text{x}+\text{C}$ is the solution to the given differential equation.
View full question & answer→Question 115 Marks
Solve the following initial value problems:
$\Big\{\text{x}\sin^2\Big(\frac{\text{y}}{\text{x}}\Big)-\text{y}\Big\}\text{dx + x dy}=0,\text{y}(1)=\frac{\pi}4$
Answer$\Big\{\text{x}\sin^2\Big(\frac{\text{y}}{\text{x}}\Big)-\text{y}\Big\}\text{dx + x dy}=0,\text{y}(1)=\frac{\pi}4$
It is a homogeneous equation. so, we put y = vx
$\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}}$
So, $\text{v + x}\frac{\text{dv}}{\text{dx}}=-\sin^2\Big(\frac{\text{vx}}{\text{x}}\Big)+\frac{\text{vx}}{\text{x}}$
$\text{x}\frac{\text{dv}}{\text{dx}}=-\sin^2\text{v}$
$\frac{\text{dv}}{\sin^2\text{v}}=-\frac{\text{dx}}{\text{x}}$
integrating both sides, we get
$\cot\Big(\frac{\text{y}}{\text{x}}\Big)=\log|\text{Cx}|$
Putting the values of x = 1 and $\text{y}=\frac{\pi}4$
$\cot\Big(\frac{\pi}{4}\Big)=\log\text{C}$
$1=\log\text{C}$
$\text{C}=\text{e}$
Hence, $\cot\Big(\frac{\text{y}}{\text{x}}\Big)=\log(\text{ex})$
View full question & answer→Question 125 Marks
Solve the following differential equations:
$\sqrt{1+\text{x}^2+\text{y}^2+\text{x}^2\text{y}^2}+\text{xy}\ \frac{\text{dy}}{\text{dx}}=0$
AnswerWe have,$\sqrt{1+\text{x}^2+\text{y}^2+\text{x}^2\text{y}^2}+\text{xy}\ \frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\sqrt{(1+\text{x}^2)(1+\text{y}^2)}+\text{xy}\ \frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\text{xy}\ \frac{\text{dy}}{\text{dx}}=-\sqrt{(1+\text{x}^2)(1+\text{y}^2)}$
$\Rightarrow\text{xy}\frac{\text{dy}}{\text{dx}}=-\sqrt{(1+\text{x}^2)}\sqrt{1+\text{y}^2)}$
$\Rightarrow\frac{\text{y}}{\sqrt{(1+\text{y}^2)}}\ \text{dy}=-\int\frac{\sqrt{(1+\text{x}^2)}}{\text{x}}\ \text{dx}$
$\Rightarrow\frac{\text{y}}{\sqrt{(1+\text{y}^2)}}\ \text{dy}=-\int\frac{\text{x}\sqrt{(1+\text{x}^2)}}{\text{x}^2}\ \text{dx}$
Putting $1+y^2=t$ and $1+x^2=u^2 \Rightarrow 2 y d y=d t$ and 2x dx = 2udu$\Rightarrow\text{y dy}=\frac{\text{dt}}{2}$
and x dx = udu$\therefore$ intregals become,
$\frac{1}{2}\int\frac{\text{dt}}{\sqrt{\text{t}}}=-\int\frac{\text{u}\times\text{u}}{\text{u}^2-1}\ \text{du}$
$\Rightarrow\sqrt{\text{t}}=-\int\frac{\text{u}^2}{\text{u}^2-1}\ \text{du}$
$\Rightarrow\sqrt{\text{t}}=-\int1+\frac{1}{\text{u}^2-1}\ \text{du}$
$\Rightarrow\sqrt{\text{t}}-\int(1)\text{du}-\int\frac{1}{\text{u}^2-1}\ \text{du}$
$\Rightarrow\sqrt{\text{t}}=-\text{u}-\frac{1}{2}\log\Big|\frac{\text{u}-1}{\text{u}+1}\Big|+\text{C}$
$\Rightarrow\sqrt{1+\text{y}^2}=-\sqrt{1+\text{x}^2}-\frac{1}{2}\log\Big|\frac{\sqrt{1+\text{x}^2}-1}{\sqrt{1+\text{x}^2}+1}\Big|+\text{C}$
$\Rightarrow\sqrt{1+\text{y}^2}+\sqrt{1+\text{x}^2}+\frac{1}{2}\log\Big|\frac{\sqrt{1+\text{x}^2}-1}{\sqrt{1+\text{x}^2}+1}\Big|=\text{C}$
View full question & answer→Question 135 Marks
Solve the following differential equation:
$\text{x dy}=(2\text{y}+2\text{x}^4+\text{x}^2)\text{dx}$
AnswerHere, $\text{x dy}=(2\text{y}+2\text{x}^4+\text{x}^2)\text{dx}$
$\text{x}\frac{\text{dy}}{\text{dx}}=2\text{y}+2\text{x}^4+\text{x}^2$
$\frac{\text{dy}}{\text{dx}}-\frac{2}{\text{x}}\text{y}=2\text{x}^3+\text{x}$
It is a linear differential equation. Comparing it with equation,
$\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
$\text{P}=-\frac{2}{\text{x}},\text{Q}=2\text{x}^3+\text{x}$
I.F. $=\text{e}^{\int\text{Pdx}}$
$=\text{e}^{-2\int\frac{1}{1+\text{x}}\text{dx}}$
$=\text{e}^{-2\log|\text{x}|}$
$=\text{e}^{\log\big(\frac{1}{\text{x}^2}\big)}$
$=\frac{1}{\text{x}^2}$
Solution of the equation is given by,
$\text{y}\times(\text{I.F.})=\int\text{Q}\times(\text{I.F.})\text{dx + C}$
$\text{y}\Big(\frac{1}{\text{x}^2}\Big)=\int\big(2\text{x}^3+\text{x}\big)\Big(\frac{1}{\text{x}^2}\Big)\text{dx + C}$
$\frac{\text{y}}{\text{x}^2}=\int\Big(2\text{x}+\frac{1}{\text{x}}\Big)\text{dx + C}$
$\frac{\text{y}}{\text{x}^2}=2\frac{\text{x}^2}2+\log|\text{x}|+\text{C}$
$\text{y}=\text{x}^4+\text{x}^2\log|\text{x}|+\text{Cx}^2$
View full question & answer→Question 145 Marks
Solve the following differential equation
$\frac{\text{dy}}{\text{dx}}=\frac{1+\text{y}^2}{\text{y}^3}$
AnswerWe have
$\frac{\text{dy}}{\text{dx}}=\frac{1+\text{y}^2}{\text{y}^3}$
$\Rightarrow\frac{\text{y}^3}{1+\text{y}^2}$
$\Rightarrow\text{dx}=\frac{\text{y}^3}{1+\text{y}^2}\ \text{dy}$
Integrating both sides, we get
$\int\text{dx}=\int\frac{\text{y}^3}{1+\text{y}^2}\text{dy}$
$\Rightarrow\text{x}=\int\frac{\text{y}+\text{y}^3-\text{y}}{1+\text{y}^2}\text{dy}$
$\Rightarrow\text{x}=\int\frac{(1+\text{y}^2)\text{y}-\text{y}}{1+\text{y}^2}\text{dy}$
$\Rightarrow\text{x}=\int\text{y dy}-\int\frac{\text{y}}{1+\text{y}^2}\ \text{dy}$
$\Rightarrow\text{x}=\frac{\text{y}^2}{2}-\int\frac{\text{y}}{1+\text{y}^2}\ \text{dy}$
Putting $1 + y^2= t $ we get
$2y\ dy = dt$
$\therefore\text{x}=\frac{\text{y}^2}{2}-\frac{1}{2}\int\frac{1}{\text{t}}\text{dt}$
$\Rightarrow\text{x}=\frac{\text{y}^2}{2}-\frac{1}{2}\log|\text{t}|+\text{C}$
$\Rightarrow\text{x}=\frac{\text{y}^2}{2}-\frac{1}{2}\log|1+\text{y}^2|+\text{C}$
Hence, $\text{x}=\frac{\text{y}^2}{2}-\frac{1}{2}\log|1+\text{y}^2|+\text{C}$ is the required solution.
View full question & answer→Question 155 Marks
Solve the following initial value problems:
$(\text{y}^4-2\text{x}^3\text{y})\text{dx}+(\text{x}^4-2\text{xy}^3)\text{dy}=0,\text{y}(1)=1$
Answer$(\text{y}^4-2\text{x}^3\text{y})\text{dx}+(\text{x}^4-2\text{xy}^3)\text{dy}=0,\text{y}(1)=1$
This is a homogeneous equation, put y = vx
$\big(\text{v}^4\text{x}^4-2\text{vx}^4\big)+\big(\text{x}^4-2\text{v}^3\text{x}^4\big)\Big[\text{v + x}\frac{\text{dv}}{\text{dx}}\Big]=0$
$\big(\text{v}^4\text{x}^4-2\text{vx}^4\big)=\big(2\text{v}^3\text{x}^4-\text{x}^4\big)\Big[\text{v + x}\frac{\text{dv}}{\text{dx}}\Big]$
$\text{vx}^4(\text{v}^3-2)=\text{x}^4(2\text{v}^3-1)\Big[\text{v + x}\frac{\text{dv}}{\text{dx}}\Big]$
$\text{v}(\text{v}^3-2)=(2\text{v}^3-1)\text{v + x}(2\text{v}^3-1)\frac{\text{dv}}{\text{dx}}$
$\text{v}\big[\text{v}^3-2-2\text{v}^3+1\big]=\text{x}(2\text{v}^3-1)\frac{\text{dv}}{\text{dx}}$
$\text{v}(-1-\text{v}^3)=\text{x}(2\text{v}^3-1)\frac{\text{dv}}{\text{dx}}$
$\text{v}(1+\text{v}^3)=\text{x}(1-2\text{v}^3)\frac{\text{dv}}{\text{dx}}$
$\frac{\text{dx}}{\text{x}}=\frac{(1-2\text{v}^3)}{\text{v}(1+\text{v}^3)}\text{dv}$
On integrating both side of the equation we get,
$\int\frac{\text{dx}}{\text{x}}=\int\frac{(1-2\text{v}^3)}{\text{v}(1+\text{v}^3)}\text{dv}$
$\Rightarrow\ \log_{\text{e}}\text{x}=\int\frac{1+\text{v}^3-3\text{v}^3}{\text{v}(1+\text{v}^3)}\text{dv}$
$\Rightarrow\ \log_{\text{e}}\text{x}=\int\frac{1+\text{v}^3}{\text{v}(1+\text{v}^3)}\text{dv}-\int\frac{3\text{v}}{\text{v}(1+\text{v}^3)}\text{dv}$
$\Rightarrow\ \log_{\text{e}}\text{x}=\int\frac{1}{\text{v}}\text{dv}-\int\frac{3\text{v}^2}{(1+\text{v}^3)}\text{dv}$
$\Rightarrow\ \log_{\text{e}}\text{x}=\log_{\text{e}}\text{v}-\int\frac{\text{dt}}{\text{t}}$
$\Rightarrow\ \log_{\text{e}}\text{x}=\log_{\text{e}}\text{v}-\log_{\text{e}}(1+\text{v}^3)+\text{C}$ let $(1+\text{v}^3)=\text{t},3\text{v}^2\text{dv}=\text{dt}$
$\Rightarrow\ \log_{\text{e}}\text{x}=\log_{\text{e}}\frac{\text{v}}{1+\text{v}^3}+\text{C}$
As $\text{v}=\frac{\text{y}}{\text{x}}$
$\Rightarrow\ \log_{\text{e}}\text{x}=\log_{\text{e}}\frac{\frac{\text{y}}{\text{x}}}{1+\text{y}^{\frac{3}{\text{x}}}}+\text{C}$
$\Rightarrow\ \log_{\text{e}}\text{x}=\log_{\text{e}}\frac{\text{yx}^2}{\text{x}^3+\text{y}^3}+\text{C}$
As y(1) = 1
$\Rightarrow\ \log_{\text{e}}1=\log_{\text{e}}\frac{1}{1+1}+\text{C}$
$\Rightarrow\ 0=\log_{\text{e}}\frac{1}2+\text{C}$
$\text{C}=-\log_{\text{e}}\frac{1}2$
$\Rightarrow\ \text{C}=\log_{\text{e}}2$
$\therefore\ \log_{\text{e}}\text{x}=\log_{\text{e}}\frac{\text{yx}^2}{\text{x}^3+\text{y}^3}+\log_{\text{e}}2$
View full question & answer→Question 165 Marks
Find one-parameter families of solution curves of the following differential equation: (or solve the following differential equation)$\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}=(\text{x}+1)\text{e}^{-\text{x}}$
AnswerWe have,
$\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}=(\text{x}+1)\text{e}^{-\text{x}}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}-\frac{1}{\text{x}}\text{y}=\Big(\frac{\text{x}+1}{\text{x}}\Big)\text{e}^{-\text{x}}\ \dots(1)$
Clearly, it is a linear differential equation of the form
$\frac{\text{dy}}{\text{dx}}+\text{Py = Q}$
where
$\text{P}=-\frac{1}{\text{x}}$
$\text{Q}=\Big(\frac{\text{x}+1}{\text{x}}\Big)\text{e}^{-\text{x}}$
$\therefore$ I.F. $=\text{e}^{\int\text{Pdx}}$
$=\text{e}^{-\int\frac{1}{\text{x}}\text{dx}}$
$=\text{e}^{-\log\text{x}}$
$=\frac{1}{\text{x}}$
Multiplying both sides of (1) by $\frac{1}{\text{x}},$ we get
$\frac{1}{\text{x}}\Big(\frac{\text{dy}}{\text{dx}}-\frac{1}{\text{x}}\text{y}\Big)=\frac{1}{\text{x}}\Big(\frac{\text{x}+1}{\text{x}}\Big)\text{e}^{-\text{x}}$
$\Rightarrow\ \frac{1}{\text{x}}\frac{\text{dy}}{\text{dx}}-\frac{1}{\text{x}^2}\text{y}=\Big(\frac{\text{x}+1}{\text{x}^2}\Big)\text{e}^{-\text{x}}$
Integrating both sides with respect to x, we get
$\frac{1}{\text{x}}\text{y}=\int\Big(\frac{1}{\text{x}}+\frac{1}{\text{x}^2}\Big)\text{e}^{-\text{x}}\text{dx + C}\ \dots(2)$
Putting $\frac{1}{\text{x}}\text{e}^{-\text{x}}=\text{t}$
$\Rightarrow\ \Big(-\frac{1}{\text{x}}\text{e}^{-\text{x}}-\frac{1}{\text{x}^2}\text{e}^{-\text{x}}\Big)\text{dx = dt}$
$\Rightarrow\ \Big(\frac{1}{\text{x}}+\frac{1}{\text{x}^2}\Big)\text{e}^{-\text{x}}\text{dx}=-\text{dt}$
Therefore (2) becomes
$\frac{1}{\text{x}}\text{y}=-\int\text{dt + C}$
$\Rightarrow\frac{1}{\text{x}}\text{y}=-\text{t + C}$
$\Rightarrow\ \frac{1}{\text{x}}\text{y}=-\frac{1}{\text{x}}\text{e}^{-\text{x}}+\text{C}$
$\Rightarrow\ \text{y}=-\text{e}^{-\text{x}}+\text{Cx}$
Hence, $\text{y}=-\text{e}^{-\text{x}}+\text{Cx}$ is the required solution.
View full question & answer→Question 175 Marks
Solve the following differential equation:
$\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}=2\sqrt{\text{y}^2-\text{x}^2}$
AnswerWe have,
$\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}=2\sqrt{\text{y}^2-\text{x}^2}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{2\sqrt{\text{y}^2-\text{x}^2}+\text{y}}{\text{x}}$
This is a homogeneous differential equation.
Putting x = vy and $\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}}$, we get
$\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{2\sqrt{\text{v}^2\text{x}^2-\text{x}^2}+\text{vx}}{\text{x}}$
$\Rightarrow\ \text{v + x}\frac{\text{dv}}{\text{dx}}=2\sqrt{\text{v}^2-1}+\text{v}$
$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=2\sqrt{\text{v}^2-1}+\text{v}-\text{v}$
$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=2\sqrt{\text{v}^2-1}$
$\Rightarrow\ \frac{1}{2\sqrt{\text{v}^2-1}}\text{dv}=\frac{1}{\text{x}}\text{dx}$
Integrating both sides, we get
$\int\frac{1}{2\sqrt{\text{v}^2-1}}\text{dv}=\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\ \int\frac{1}{2\sqrt{\text{v}^2-1}}\text{dv}=2\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\ \log\Big|\text{v}+\sqrt{\text{v}^2-1}\Big|=2\log|\text{x}|+\log\text{C}$
$\Rightarrow\ \text{v}+\sqrt{\text{v}^2-1}=\text{Cx}^2$
Putting $\text{v}=\frac{\text{y}}{\text{x}}$, we get
$\therefore\ \frac{\text{y}}{\text{x}}+\sqrt{\frac{\text{y}^2}{\text{x}^2}-1}=\text{Cx}^2$
$\Rightarrow\ \text{y}+\sqrt{\text{y}^2-\text{x}^2}=\text{Cx}^3$
Hence, $\text{y}+\sqrt{\text{y}^2-\text{x}^2}=\text{Cx}^3$ is the required solution.
View full question & answer→Question 185 Marks
Find one-parameter families of solution curves of the following differential equation: (or solve the following differential equation)$(\text{x + y})\frac{\text{dy}}{\text{dx}}=1$
AnswerWe have,
$(\text{x + y})\frac{\text{dy}}{\text{dx}}=1$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x + y}}$
$\Rightarrow\ \frac{\text{dx}}{\text{dy}}=\text{x + y}$
$\Rightarrow\ \frac{\text{dx}}{\text{dy}}-\text{x}=\text{y}\ \dots(1)$
Clearly, it is a linear differential equation of the form
$\frac{\text{dx}}{\text{dy}}+\text{Px = Q}$
where
$\text{P}=-1$
$\text{Q}=\text{y}$
$\therefore$ I.F. $=\text{e}^{\int\text{Pdx}}$
$=\text{e}^{\int-1\text{dy}}$
$=\text{e}^{-\text{y}}$
Multiplying both sides of (1) by $\text{e}^{-\text{y}},$ we get
$\text{e}^{-\text{y}}\Big(\frac{\text{dx}}{\text{dy}}-\text{x}\Big)=\text{e}^{-\text{y}}\text{y}$
$\Rightarrow\ \text{e}^{-\text{y}}\frac{\text{dx}}{\text{dy}}-\text{e}^{-\text{y}}\text{x}=\text{e}^{-\text{y}}\text{y}$
Integrating both sides with respect to x, we get
$\text{e}^{-\text{y}}\text{x}=\int\text{ye}^{-\text{y}}\text{dy + C}$
$\Rightarrow\ \text{e}^{-\text{y}}\text{x}=\text{y}\int\text{e}^{-\text{y}}\text{dy}-\int\Big[\frac{\text{d}}{\text{dy}}(\text{y})\int\text{e}^{-\text{y}}\text{dy}\Big]\text{dy + C}$
$\Rightarrow\ \text{e}^{-\text{y}}\text{x}=-\text{ye}^{-\text{y}}-\text{e}^{-\text{y}}+\text{C}$
$\Rightarrow\ \text{e}^{-\text{y}}\text{x}+\text{ye}^{-\text{y}}+\text{e}^{-\text{y}}=\text{C}$
$\Rightarrow\ (\text{x + y}+1)\text{e}^{-\text{y}}=\text{C}$
$\Rightarrow\ (\text{x + y}+1)=\text{Ce}^{\text{y}}$
Hence, $(\text{x + y}+1)=\text{Ce}^{\text{y}}$ is the required solution.
View full question & answer→Question 195 Marks
Find the particular solution of the differential equation $\frac{\text{dx}}{\text{dy}} + \text{x}\cot \text{y}=2\text{y} + \text{y}^{2} \cot \text{y},\text{ y}\neq0$ given that x = 0 when $\text{y}=\frac{\pi}{2}$
AnswerWe have,
$\frac{\text{dx}}{\text{dy}} + \text{x}\cot \text{y}=2\text{y} + \text{y}^{2} \cot \text{y}\ ...(1)$
Clearly, it is a linear differential equation of the form
$\frac{\text{dx}}{\text{dy}}+\text{Px}=\text{Q}$
Where $\text{P}=\cot\text{y}$ and $\text{Q}=2\text{y} + \text{y}^{2} \cot \text{y}$
$\therefore\ \text{I}.\text{F}.=\text{e}^{\int\text{P}\text{dy}}$
$=\text{e}^{\int\cot\text{y}\text{ dy}}$
$=\text{e}^{\log|\sin\text{y}|}$
$=\sin\text{y}$
Multiplying both sides of (1) by $\text{I.F.}=\sin\text{y},$ we get
$\sin\text{y}\Big(\frac{\text{dx}}{\text{dy}}+\text{x}\cot\text{y}\Big)=\sin\text{y}\big(\text{y}^2\cot\text{y} + 2\text{y}\big)$
$\Rightarrow\sin \text{y}\frac{\text{dx}}{\text{dy}}+\text{x}\cos\text{y}=\text{y}^2\cos\text{y}+2\text{y}\sin\text{y}$
Integrating both sides with respect to y, we get
$\text{x}\sin \text{y}=\int\text{y}^{2}\cos\text{y}\text{ dy}+\int2\text{y}\sin\text{y}\text{ dy}+\text{C}$
$\Rightarrow\text{x}\sin\text{y}=\text{y}^2\int\cos\text{y dy}-\int\Big[\frac{\text{d}}{\text{dy}}(\text{y}^2)\int\cos\text{y dy}\Big]\text{dy}+\int2\text{y}\sin\text{y dy}+\text{C}$
$ \Rightarrow\text{x} \sin\text{y}=\text{y}^2 \sin\text{y}-\int2\text{y}\sin\text{y} + \int2\text{y}\sin\text{y}\text{ dy} + \text{C}$
$\Rightarrow\text{x}\sin \ \text{y}=\text{y}^2 \sin \ \text{y} + \text{C}$
Now,
$\therefore 0\times\sin \frac{\pi}{2}=\frac{\pi^2}{4} \sin \frac{\pi}{2}+\text{C}$
$\Rightarrow\text{C}=-\frac{\pi^2}{4}$
Putting the value of C, we get
$\text{x}\sin \ \text{y}=\text{y}^2\sin \text{y}-\frac{\pi^2}{4}$
Hence, $\text{x}\sin \ \text{y}=\text{y}^2\sin \text{y}-\frac{\pi^2}{4}$ is the required solution.
View full question & answer→Question 205 Marks
Solve the following differential equation:
$\frac{\text{dy}}{\text{dx}}+\text{y}\cos\text{x}=\sin\text{x}\cos\text{x}$
AnswerWe have,
$\frac{\text{dy}}{\text{dx}}+\text{y}\cos\text{x}=\sin\text{x}\cos\text{x}\ \dots(1)$
Clearly, it is a linear differential equation of the form
$\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
where
$\text{P}=\cos\text{x}$
$\text{Q}=\sin\text{x}\cos\text{x}$
$\therefore$ I.F. $=\text{e}^{\int\text{Pdx}}$
$=\text{e}^{\int\cos\text{xdx}}$
$=\text{e}^{\sin\text{x}}$
Multiplying both sides of (1) by $\text{e}^{\sin\text{x}}$ we get
$\text{e}^{\sin\text{x}}\Big(\frac{\text{dy}}{\text{dx}}+\text{y}\cos\text{x}\Big)=\text{e}^{\sin\text{x}}\sin\text{x}\cos\text{x}$
$\Rightarrow\ \text{e}^{\sin\text{x}}\frac{\text{dy}}{\text{dx}}+\text{e}^{\sin\text{x}}\text{y}\cos\text{x}=\text{e}^{\sin\text{x}}\sin\text{x}\cos\text{x}$
Integrating both sides with respect to x, we get
$\text{y}\text{e}^{\sin\text{x}}=\int\text{e}^{\sin\text{x}}\sin\text{x}\cos\text{x dx + C}$
$\Rightarrow\ \text{y}\text{e}^{\sin\text{x}}=\text{I + C}\ \dots(2)$
where
$\text{I}=\int\text{e}^{\sin\text{x}}\sin\text{x}\cos\text{x dx}$
Putting $\text{t}=\sin\text{x},$ we get
$\text{dt}=\cos\text{x dx}$
$\therefore\ \text{I}=\int\text{e}^{\text{t}}\text{t dt}$
$=\text{t}\int\text{e}^{\text{t}}\text{dt}-\int\Big[\frac{\text{d}}{\text{dt}}(\text{t})\int\text{e}^{\text{t}}\text{dt}\Big]\text{dt}$
$=\text{te}^{\text{t}}-\text{e}^{\text{t}}$
$=\text{e}^{\text{t}}(\text{t}-1)$
$=\text{e}^{\sin\text{x}}(\sin\text{x}-1)$
Putting the value of I in (2), we get
$\text{ye}^{\sin\text{x}}=\text{e}^{\sin\text{x}}(\sin\text{x}-1)+\text{C}$
$\Rightarrow\ \text{y}=\sin\text{x}-1+\text{Ce}^{-\sin\text{x}}$
Hence, $\text{y}=\sin\text{x}-1+\text{Ce}^{-\sin\text{x}}$ is the required solution.
View full question & answer→Question 215 Marks
Find one-parameter families of solution curves of the following differential equation: (or solve the following differential equation)$\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}=\text{x}^4$
AnswerWe have,
$\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}=\text{x}^4$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}+\frac{1}{\text{x}}\text{y}=\text{x}^3\ \dots(1)$
Clearly, it is a linear differential equation of the form
$\frac{\text{dy}}{\text{dx}}+\text{Py = Q}$
where
$\text{P}=\frac{1}{\text{x}}$
$\text{Q}=\text{x}^3$
$\therefore$ I.F. $=\text{e}^{\int\text{Pdx}}$
$=\text{e}^{\int\frac{1}{\text{x}}\text{dx}}$
$=\text{e}^{\log\text{x}}$
$={\text{x}}$
Multiplying both sides of (1) by x, we get
${\text{x}}\Big(\frac{\text{dy}}{\text{dx}}+\frac{1}{\text{x}}\text{y}\Big)=\text{x}.\text{x}^3$
$\Rightarrow\ {\text{x}}\frac{\text{dy}}{\text{dx}}+\text{x}=\text{x}^4$
Integrating both sides with respect to x, we get
$\text{xy}=\int\text{x}^4\text{dx + C}$
$\Rightarrow\ \text{xy}=\frac{\text{x}^5}{5}+\text{C}$
$\Rightarrow\ \text{y}=\frac{\text{x}^4}{5}+\frac{\text{C}}{\text{x}}$
Hence, $\text{y}=\frac{\text{x}^4}{5}+\frac{\text{C}}{\text{x}}$ is the required solution.
View full question & answer→Question 225 Marks
Solve the following differential equation $(\text{x}+2\text{y}^2)\frac{\text{dy}}{\text{dx}}=\text{y},$ given that when x = 2, y = 1.
AnswerWe have, $(\text{x}+2\text{y}^2)\frac{\text{dy}}{\text{dx}}=\text{y}$ $\Rightarrow\ \frac{\text{dx}}{\text{dy}}=\frac{1}{\text{y}}(\text{x}+2\text{y}^2)$ $\Rightarrow\ \frac{\text{dx}}{\text{dy}}-\frac{1}{\text{y}}\text{x}=2\text{y}\ \dots(1)$ Clearly, it is a linear differential equation of the form $\frac{\text{dx}}{\text{dy}}+\text{Px = Q}$ where $\text{P}=-\frac{1}{\text{y}}$ $\text{Q}=2\text{y}$ $\therefore$ I.F. $=\text{e}^{\int\text{Pdy}}$ $=\text{e}^{-\int\frac{1}{\text{y}}\text{dy}}$ $=\text{e}^{-\log\text{y}}$ $=\frac{1}{\text{y}}$Multiplying both sides of (1) by $\frac{1}{\text{y}},$ we get
$\frac{1}{\text{y}}\Big(\frac{\text{dx}}{\text{dy}}-\frac{1}{\text{y}}\text{x}\Big)=\frac{1}{\text{y}}\times2\text{y}$ $\Rightarrow\ \frac{1}{\text{y}}\frac{\text{dx}}{\text{dy}}-\frac{1}{\text{y}^2}\text{x}=2$ Integrating both sides with respect to y, we get $\text{x}\frac{1}{\text{y}}=\int2\text{dy + C}$ $\Rightarrow\ \text{x}\frac{1}{\text{y}}=2\text{y + C}$ $\Rightarrow\ \text{x}=2\text{y}^2+\text{Cy}\ \dots(2)$ Now, $\text{y}=1$ at $\text{x}=2$ $\therefore\ 2=2+\text{C}$ $\Rightarrow\ \text{C}=0$ Putting the value of C in (2), we get $\text{x}=2\text{y}^2$ Hence, $\text{x}=2\text{y}^2$ is the required solution.
View full question & answer→Question 235 Marks
Form the differential equation of the family of curves represented by the equation (a being the perimeter):$(2\text{x}+\text{a})^2+\text{y}^2=\text{a}^2$
AnswerThe equation of the family of curves is
$(2\text{x}+\text{a})^2+\text{y}^2=\text{a}^2\ ...(1)$
where a is a parameter.
As this equation has only one parameter, we shall get a differential equation of first order.
Differentiating (1) with respect to x, we get
$2(2\text{x}+\text{a})\times2+2\text{y}\frac{\text{dy}}{\text{dx}}=0\ ...(2)$
Now, from (1), we get
$4\text{x}^2+4\text{ax}+\text{a}^2+\text{y}^2=\text{a}^2$
$\Rightarrow4\text{ax}=-\text{y}^2-4\text{x}^2$
$\Rightarrow\text{a}=-\frac{(4\text{x}^2+\text{y}^2)}{4\text{x}}$
putting the value of a in (2), we get
$4\Big(2\text{x}-\frac{4\text{x}^2+\text{y}^2}{4\text{x}}\Big)+2\text{y}\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow4\Big(\frac{8\text{x}^2-4\text{x}^2-\text{y}^2}{4\text{x}}\Big)+2\text{y}\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow4\text{x}^2-\text{y}^2+2\text{xy}\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\text{y}^2-4\text{x}^2-2\text{xy}\frac{\text{dy}}{\text{dx}}=0$
It is the required differential equation.
View full question & answer→Question 245 Marks
Solve the following differential equation
$\frac{\text{dy}}{\text{dx}}-\text{x}\sin^2\text{x}=\frac{1}{\text{x}\log\text{x}}$
Answer$\frac{\text{dy}}{\text{dx}}-\text{x}\sin^2\text{x}=\frac{1}{\text{x}\log\text{x}}$
$\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}\log\text{x}}+\text{x}\sin^2\text{x}$
$\text{dy}=\Big(\frac{1}{\text{x}\log\text{x}}+\text{x}\sin^2\text{x}\Big)\text{dx}$
$\int\text{dy}=\int\frac{1}{\text{x}\log\text{x}}\text{dx}+\int\text{x}\sin^2\text{x dx}$
$\text{y}=\text{I}_1+\text{I}_2$
$\text{I}_1=\int\frac{1}{\text{x}\log\text{x}}\text{ dx}$
Let $\log\text{x}=\text{t}$
$\frac{1}{\text{x}}\ \text{dx}=\text{dt}$
$\text{I}_1=\int\frac{\text{dt}}{\text{t}}$
$=\log|\text{t}|+\text{C}_1$
$\text{I}_1=\log|\log\text{x}|+\text{C}_1$
$\text{I}_2=\int\text{x}\sin^2\text{x dx}$
$=\int\text{x}\frac{(1-\cos2\text{x})}{2}\ \text{dx}$
$=\frac{1}{2}\int(\text{x}-\text{x}\cos2\text{x})\text{dx}$
$=\frac{1}{2}\int\text{x dx}-\frac{1}{2}\int\text{x}\cos2\text{x dx}$
$=\frac{1}{2}\Big(\frac{\text{x}^2}{2}\Big)-\frac{1}{2}[\text{x}\int\cos2\text{x dx}-\int(1\times\int\cos2\text{x dx})\text{dx}]+\text{C}_2$
$=\frac{\text{x}^2}{4}-\frac{1}{2}\Big[\frac{\text{x}\sin\text{x}}{2}+\frac{\cos2\text{x}}{4}\Big]+\text{C}_2$
$\text{I}_2=\frac{\text{x}^2}{4}-\frac{\text{x}\sin2\text{x}}{4}-\frac{\cos2\text{x}}{8}+\text{C}_2$
Put the values of $l_1$ and $l_2$ in equation (1)
$\text{y}=\text{I}_1+\text{I}_2$
$\text{y}=\log|\log\text{x}|+\frac{\text{x}^2}{4}-\frac{\text{x}\sin2\text{x}}{4}-\frac{\cos2\text{x}}{8}+\text{C}\text{ as}\text{ C}_1+\text{C}_2=\text{C}$
View full question & answer→Question 255 Marks
Find one-parameter families of solution curves of the following differential equation: (or solve the following differential equation)$\frac{\text{dy}}{\text{dx}}+\text{y}\cos\text{x}=\text{e}^{\sin\text{x}}\cos\text{x}$
AnswerWe have,
$\frac{\text{dy}}{\text{dx}}+\text{y}\cos\text{x}=\text{e}^{\sin\text{x}}\cos\text{x}\ \dots(1)$
Clearly, it is a linear differential equation of the form
$\frac{\text{dy}}{\text{dx}}+\text{Py = Q}$
where
$\text{P}=\cos\text{x}$
$\text{Q}=\text{e}^{\sin\text{x}}\cos\text{x}$
$\therefore$ I.F. $=\text{e}^{\int\text{Pdx}}$
$=\text{e}^{\int\cos\text{xdx}}$
$=\text{e}^{\sin\text{x}}$
Multiplying both sides of (1) by $\text{e}^{\sin\text{x}},$ we get
$\text{e}^{\sin\text{x}}\Big(\frac{\text{dy}}{\text{dx}}+\text{y}\cos\text{x}\Big)=\text{e}^{\sin\text{x}}\times\text{e}^{\sin\text{x}}\cos\text{x}$
$\Rightarrow\ \text{e}^{\sin\text{x}}\frac{\text{dy}}{\text{dx}}+\text{y}\text{e}^{\sin\text{x}}\cos\text{x}=\text{e}^{2\sin\text{x}}\cos\text{x}$
Integrating both sides with respect to x, we get
$\text{e}^{\sin\text{x}}\text{y}=\int\text{e}^{\sin\text{x}}\cos \text{x}\text{dx + C}$
$\Rightarrow\ \text{e}^{\sin\text{x}}\text{y}=\text{I + C}\ \dots(2)$
where,
$\text{I}=\int\text{e}^{\sin\text{x}}\cos\text{xdx}$
Putting $\text{t}=\sin\text{x},$ we get
$\text{dt}=\cos\text{xdx}$
$\therefore\ \text{I}=\int\text{e}^{2\text{t}}\text{dt}$
$=\frac{\text{e}^{2\text{t}}}{2}$
$=\frac{\text{e}^{2\sin\text{x}}}{2}$
Putting the value of I in (2), we get
$\text{e}^{\sin\text{x}}\text{y}=\frac{\text{e}^{2\sin\text{x}}}{2}+\text{C}$
$\Rightarrow\ \text{y}=\frac{\text{e}^{\sin\text{x}}}{2}+\text{C}\text{e}^{-\sin\text{x}}$
Hence, $\text{y}=\frac{\text{e}^{\sin\text{x}}}{2}+\text{C}\text{e}^{-\sin\text{x}}$ is the required solution.
View full question & answer→Question 265 Marks
Solve the following differential equation:
$\frac{\text{dy}}{\text{dx}} = \sec(\text{x}+\text{y})$
AnswerWe have,
$\frac{\text{dy}}{\text{dx}} = \sec(\text{x}+\text{y})$
$\frac{\text{dy}}{\text{dx}} = \frac{1}{\cos(\text{x}+\text{y})}$
Let $\text{ x}+\text{y} = \text{v}$
$\Rightarrow 1+\frac{\text{dy}}{\text{dx}} = \frac{\text{dv}}{\text{dx}}$
$\Rightarrow \frac{\text{dv}}{\text{dx}} = \frac{\text{dv}}{\text{dx}}-1$
$\therefore\frac{\text{dv}}{\text{dx}}-1 = \frac{1}{\cos\text{v}}$
$\Rightarrow\frac{\text{dv}}{\text{dx}} = \frac{\cos\text{v}+1}{\cos\text{v}}$
$\Rightarrow \frac{\cos \text{v}}{\cos\text{v}+1}\text{dv} = \text{dx}$
Integrating both sides, we get
$\int \frac{\cos\text{v}}{\cos\text{v}+1}\text{dv} = \int\text{dx}$
$\Rightarrow \int\frac{\cos\text{v}(1-\cos\text{v})}{1-\cos^2\text{v}}\text{dv} = \int\text{dx}$
$\Rightarrow \int\frac{\cos\text{v}(1-\cos\text{v})}{\sin^2\text{v}}\text{dv} = \int\text{dx}$
$\Rightarrow \int\frac{\cos\text{v}-\cos^2\text{v}}{\sin^2\text{v}}\text{dv} = \int\text{dx}$
$\Rightarrow\int(\cot\text{v}\text{ cosec}\text{ v}-\cot^2\text{v})\text{dv} = \int\text{dx}$
$\Rightarrow \int(\cot\text{v}\text{ cosec }\text{v}-\text{cosec}^2\text{v}+1)\text{dv} = \int\text{dx}$
$\Rightarrow -\text{cosec }\text{v}+\cot\text{v}+\text{v} = \text{x}+\text{C}$
$\Rightarrow -\text{cosec}(\text{x}+\text{y})+\cot(\text{x}+\text{y})+\text{x}+\text{y} = \text{x}+\text{C}$
$\Rightarrow -\text{cosec}(\text{x}+\text{y})+\cot(\text{x}+\text{y})+\text{y} = \text{C}$
$\Rightarrow \frac{-1+\cos(\text{x}+\text{y})}{\sin(\text{x}+\text{y})}+\text{y}=\text{C}$
$\Rightarrow -\tan\Big(\frac{\text{x}+\text{y}}{2}\Big)+\text{y} = \text{C}$
$\Rightarrow \text{y} = \tan\Big(\frac{\text{x}+\text{y}}{2}\Big)+\text{C}$
View full question & answer→Question 275 Marks
Solve the following differential equation:
$(\text{x}+\tan\text{y})\text{dy}=\sin2\text{y dx}$
View full question & answer→Question 285 Marks
Solve the following differential equation:
$(\text{x}^2+3\text{xy}+\text{y}^2)\text{dx}-\text{x}^2\text{dy}=0$
AnswerWe have,
$(\text{x}^2+3\text{xy}+\text{y}^2)\text{dx}-\text{x}^2\text{dy}=0$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{\text{x}^2+3\text{xy}+\text{y}^2}{\text{x}^2}$
This is a homogeneous differential equation.
Putting x = vy and $\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}}$, we get
$\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{\text{x}^2+3\text{vx}^2+\text{v}^2\text{x}^2}{\text{x}^2}$
$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=1+3\text{v}+\text{v}^2-\text{v}$
$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=1+\text{v}^2+2\text{v}$
$\Rightarrow\ \frac{1}{1+\text{v}^2+2\text{v}}\text{dv}=\frac{1}{\text{x}}\text{dx}$
Integrating both sides, we get
$\int\frac{1}{1+\text{v}^2+2\text{v}}\text{dv}=\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\ \int\frac{1}{(1+\text{v})^2}\text{dv}=\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\ -\frac{1}{(1+\text{v})}=\log|\text{x}|+\text{C}$
$\Rightarrow\ \log|\text{x}|+\frac{1}{(1+\text{v})}=-\text{C}$
Putting $\text{v}=\frac{\text{y}}{\text{x}}$, we get
$\therefore\ \log|\text{x}|+\frac{\text{x}}{(\text{x + y})}=\text{C}_1$
where
$C_1= -C$
Hence, $\log|\text{x}|+\frac{\text{x}}{(\text{x + y})}=\text{C}_1$ is the required solution.
View full question & answer→Question 295 Marks
Solve the following initial value problems:
$\frac{\text{dy}}{\text{dx}}=\frac{\text{y}(\text{x}+2\text{y})}{\text{x}(2\text{x}+\text{y})},\text{y}(1)=2$
Answer$\frac{\text{dy}}{\text{dx}}=\frac{\text{y}(\text{x}+2\text{y})}{\text{x}(2\text{x}+\text{y})},\text{y}(1)=2$
This is a homogeneous equation, put y = vx
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}}$
$\Rightarrow\ \text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{\text{v}(\text{x}+2\text{vx})}{(2\text{x + vx})}$
$\Rightarrow\ \text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{\text{v}(1+2\text{v})}{(2+\text{v})}$
$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{\text{v}+2\text{v}^2-2\text{v}-\text{v}^2}{(2+\text{v})}$
$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{\text{v}^2-\text{v}}{(2+\text{v})}$
$\Rightarrow\ \frac{(2+\text{v})\text{dv}}{(\text{v}^2-\text{v})}=\frac{\text{dx}}{\text{x}}$
On integrating both sides of the equation we get,
$\int\frac{2+\text{v}}{(\text{v}^2-\text{v})}\text{dv}=\int\frac{\text{dx}}{\text{x}}$
$\Rightarrow\ \int\frac{2}{\text{v}(\text{v}-1)}\text{dv}+\int\frac{\text{v}}{\text{v}(\text{v}-1)}\text{dv}=\int\frac{\text{dx}}{\text{x}}$
$\Rightarrow\ 2\Big[\int\frac{1}{(1-\text{v})}\text{dv}-\int\frac{1}{\text{v}}\text{dv}\Big]+\int\frac{1}{\text{v}-1}\text{dv}=\log_{\text{e}}\text{x + C}$
$\Rightarrow\ 2\big[\log_{\text{e}}(\text{v}-1)-\log_{\text{e}}\text{v}\big]+\log_{\text{e}}(\text{v}-1)=\log_{\text{e}}\text{x + C}$
$2\Big[\log_{\text{e}}\Big(\frac{\text{v}-1}{\text{v}}\Big)\Big]+\log_{\text{e}}(\text{v}-1)=\log_{\text{e}}\text{x + C}$
$2\log_{\text{e}}\Big(\frac{\text{y}-\text{x}}{\text{y}}\Big)+\log_{\text{e}}\Big(\frac{\text{y}-\text{x}}{\text{x}}\Big)=\log_{\text{e}}\text{x + C}$
As y(1) = 2
$2\log_{\text{e}}\Big(\frac{2-1}{2}\Big)+\log_{\text{e}}\Big(\frac{2-1}{1}\Big)=\log_{\text{e}}1+\text{C}$
$2\log_{\text{e}}\frac{1}2+\log_{\text{e}}1=\log_{\text{e}}1+\text{C}$
$-2\log_{\text{e}}2+0=0+\text{C}$
$-2\log_{\text{e}}2=\text{C}$
$\therefore\ 2\log_{\text{e}}\Big(\frac{\text{y}-\text{x}}{\text{y}}\Big)+\log_{\text{e}}\Big(\frac{\text{y}-\text{x}}{\text{x}}\Big)=\log_{\text{e}}\text{x}-2\log_{\text{e}}2$
View full question & answer→Question 305 Marks
The rate of increase in the number of bacteria in a certain bacteria culture is proportional to the number present. Given the number triples in 5 hrs, find how many bacteria will be present after 10 hours. Also find the time necessary for the number of bacteria to be 10 times the number of initial present.
AnswerLet A be the amount of bacteria present at time t and A be the initial amount of bacteria. Here,
$\frac{\text{dA}}{\text{dt}}\propto\text{A}$
$\frac{\text{dA}}{\text{dt}}=\lambda\text{A}$
$\int \frac{\text{dA}}{\text{A}}=\int\lambda\text{dt}$
$\log\text{A}=\lambda\text{t}+\text{C}\ ...(\text{i})$
When $t = 0, A = A0$
$\log(\text{A}_{0})=0+\text{C}$
$\text{C}=\log\text{A}_{0}$
Using equation (i)
$\log\text{A}=\lambda\text{t}+\log\text{A}_{0}$
$\log\big(\frac{\text{A}}{\text{A}_{0}}\big)=\lambda\text{t}\ ...(\text{ii})$
Given, bacteria triples is 5 hours, so
$\log\big(\frac{3\text{A}_{0}}{\text{A}_{0}}\big)=5\lambda$
$\log3=5\lambda$
$\lambda=\frac{\log3}{5}$
Putting the values of eq. (ii)
$\log\big(\frac{\text{A}}{\text{A}_{0}}\big)=\frac{\log3}{5}\text{t}$
Case I = let $A_1$ be the number of 10 hours,
$\log\big(\frac{\text{A}_1}{\text{A}_{0}}\big)=\frac{\log3}{5}\times10$
$\log\big(\frac{\text{A}_1}{\text{A}_{0}}\big)=2\log3$
$\log\big(\frac{\text{A}_1}{\text{A}_{0}}\big)=2(1.0986)$
$\log\big(\frac{\text{A}_1}{\text{A}_{0}}\big)=2.1972$
Thus, There will 9 times the present is 10 hours.
Case II = let $t_1$ be the number of 10 hours,
$\log\big(\frac{\text{A}}{\text{A}_{0}}\big)=\frac{\log3}{5}\times\text{t}_{1}$
$\log\big(\frac{\text{10A}}{\text{A}_{0}}\big)=\frac{\log3}{5}\times\text{t}_{1}$
$5\log10= \log3 \text{t}_{1}$
$\frac{5\log10}{\log3}=\text{t}_{1}$
Required time is $\frac{5\log10}{\log3}\ \text{hours}$.
View full question & answer→Question 315 Marks
Solve the following differential equation:
$(\sin\text{x})\frac{\text{dy}}{\text{dx}}+\text{y}\cos\text{x}=2\sin^2\text{x}\cos\text{x}$
AnswerWe have,
$(\sin\text{x})\frac{\text{dy}}{\text{dx}}+\text{y}\cos\text{x}=2\sin^2\text{x}\cos\text{x}$
$\frac{\text{dy}}{\text{dx}}+\text{y}\cot\text{x}=2\sin\text{x}\cos\text{x}\ \dots(1)$
Clearly, it is a linear differential equation of the form
$\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
where
$\text{P}=\cot\text{x}$
$\text{Q}=2\sin\text{x}\cos\text{x}$
$\therefore$ I.F. $=\text{e}^{\int\text{Pdx}}$
$=\text{e}^{\int\cot\text{xdx}}$
$=\text{e}^{\log|\sin\text{x}|}=\sin\text{x}$
Multiplying both sides of (1) by $\sin\text{x},$ we get
$\sin\text{x}\Big(\frac{\text{dy}}{\text{dx}}+\text{y}\cot\text{x}\Big)=\sin\text{x}\times2\sin\text{x}\cos\text{x}$
$\sin\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\cos\text{x}=2\sin^2\text{x}\cos\text{x}$
Integrating both sides with respect to x, we get
$\text{y}\sin\text{x}=2\int\sin^2\text{x}\cos\text{x dx + C}\ \dots(2)$
Putting $\sin\text{x}=\text{t}$
$\Rightarrow\ \cos\text{x dx = dt}$
Therefore, (2) becomes
$\text{y}\sin\text{x}=2\int\text{t}^2\text{dt + C}$
$\Rightarrow \text{y}\sin\text{x}=\frac{2}3\text{t}^3+\text{C}$
$\Rightarrow \text{y}\sin\text{x}=\frac{2}3\sin^3\text{x}+\text{C}$
Hence, $\text{y}\sin\text{x}=\frac{2}3\sin^3\text{x}+\text{C}$ is the required solution.
View full question & answer→Question 325 Marks
Find the particular solution of the differential equation $\frac{\text{dy}}{\text{dx}}=\frac{\text{xy}}{\text{x}^2+\text{y}^2}$ given that y = 1 when x = 0.
Answer$\frac{\text{dy}}{\text{dx}}=\frac{\text{xy}}{\text{x}^2+\text{y}^2}\ \dots(1)$
Let y = xv
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}}$
Substituting the value of y = xv and $\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}}$ in (1), we get
$\therefore\ \text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{\text{x}^2\text{v}}{\text{x}^2+\text{x}^2\text{v}^2}$
$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{\text{v}}{1+\text{v}^2}-\text{v}$
$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{-\text{v}^3}{1+\text{v}^2}$
$\Rightarrow\ \frac{1+\text{v}^2}{-\text{v}^3}\text{dv}=\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\ \int\frac{1+\text{v}^2}{-\text{v}^3}\text{dv}=\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\ \frac{1}{2\text{v}^2}-\log\text{v}=\log\text{x +C}$
$\Rightarrow\ \frac{1}{2\big(\frac{\text{y}}{\text{x}}\big)^2}-\log\frac{\text{y}}{\text{x}}=\log\text{x +C}$
$\Rightarrow\ \frac{\text{x}^2}{2\text{y}^2}-\log\frac{\text{y}}{\text{x}}=\log\text{x + C}\ \dots(2)$
$\Rightarrow\ \frac{0}2-\log\frac{1}0=\log0+\text{C}$
$\Rightarrow\ \text{C}=0$
Substituting the value of C in (2), we get
$\frac{\text{x}^2}{2\text{y}^2}-\log\frac{\text{y}}{\text{x}}=\log{\text{x}}$
$\Rightarrow\ \frac{\text{x}^2}{2\text{y}^2}=\log{\text{x}}+\log\frac{\text{y}}{\text{x}}$
$\Rightarrow\ \frac{\text{x}^2}{2\text{y}^2}=\log{\text{y}}$
View full question & answer→Question 335 Marks
Solve the following differential equation:
$\frac{\text{y}}{\text{x}}\cos\Big(\frac{\text{y}}{\text{x}}\Big)\text{dx}-\Big\{\frac{\text{x}}{\text{y}}\sin\Big(\frac{\text{y}}{\text{x}}\Big)+\cos\Big(\frac{\text{y}}{\text{x}}\Big)\Big\}\text{dy}=0$
AnswerHere, $\frac{\text{y}}{\text{x}}\cos\Big(\frac{\text{y}}{\text{x}}\Big)\text{dx}-\Big\{\frac{\text{x}}{\text{y}}\sin\Big(\frac{\text{y}}{\text{x}}\Big)+\cos\Big(\frac{\text{y}}{\text{x}}\Big)\Big\}\text{dy}=0$
$\frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{y}}{\text{x}}\cos\big(\frac{\text{y}}{\text{x}}\big)}{\frac{\text{x}}{\text{y}}\sin\big(\frac{\text{y}}{\text{x}}\big)+\cos\big(\frac{\text{y}}{\text{x}}\big)}$
It is a homogeneous equation
Put y = vx
and $\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}}$
So,
$\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{\frac{\text{vx}}{\text{x}}\cos\big(\frac{\text{vx}}{\text{x}}\big)}{\frac{\text{x}}{\text{vx}}\sin\big(\frac{\text{vx}}{\text{x}}\big)+\cos\big(\frac{\text{vx}}{\text{x}}\big)}$
$=\frac{\text{v}\cos\text{v}}{\frac{1}{\text{v}}\sin\text{v}+\cos\text{v}}$
$\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{\text{v}^2\cos\text{v}}{\sin\text{v}+\text{v}\cos\text{v}}$
$\text{x}\frac{\text{dv}}{\text{dx}}=\frac{\text{v}^2\cos\text{v}}{\sin\text{v}+\text{v}\cos\text{v}}-\text{v}$
$$$\text{x}\frac{\text{dv}}{\text{dx}}=\frac{\text{v}^2\cos\text{v}-\text{v}\sin\text{v}-\text{v}^2\cos\text{v}}{\sin\text{v}+\text{v}\cos\text{v}}$
$\text{x}\frac{\text{dv}}{\text{dx}}=\frac{-\text{v}\sin\text{v}}{\sin\text{v}+\text{v}\cos\text{v}}$
$\frac{\sin\text{v}+\text{v}\cos\text{v}}{\text{v}\sin\text{v}}\text{dv}=-\frac{\text{dx}}{\text{x}}$
$\int\Big(\frac{1}{\text{v}}+\cot\text{v}\Big)\text{dv}=-\log|\text{x}|+\log|\text{C}|$
$\log|\text{v}|+\log|\sin\text{v}|=\log\Big|\frac{\text{C}}{\text{x}}\Big|$
$\log|\text{v}\sin\text{v}|=\log\Big|\frac{\text{C}}{\text{x}}\Big|$
$|\text{v}\sin\text{v}|=\Big|\frac{\text{C}}{\text{x}}\Big|$
$\Big|\text{x}\Big(\frac{\text{y}}{\text{x}}\Big)\sin\Big(\frac{\text{y}}{\text{x}}\Big)\Big|=|\text{C}|$
$\Big|\text{y}\sin\frac{\text{y}}{\text{x}}\Big|=\text{C}$
View full question & answer→Question 345 Marks
Show that $\text{y}=\frac{\text{c}-\text{x}}{1+\text{cx}}$ is a solution of the differential equation $(1+\text{x}^2)\frac{\text{dy}}{\text{dx}}+(1+\text{y}^2)=0.$
AnswerWe have,
$\text{y}=\frac{\text{c}-\text{x}}{1+\text{cx}}\ ...(1)$
Differential both sides of (1) with respect to x, we get
$\frac{\text{dy}}{\text{dx}}=\frac{(1+\text{cx})(-1)-(\text{c}-\text{x})(\text{c})}{(1+\text{cx})^2}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{-1-\text{cx}-\text{c}^2+\text{cx}}{(1+\text{cx})^2}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1+\text{c}^2}{(1+\text{cx})^2}\ ...(2)$
Now,
$(1+\text{x}^2)\frac{\text{dy}}{\text{dx}}+(1+\text{y}^2)$
$=-(1+\text{x}^2)\frac{(1+\text{c}^2)}{(1+\text{cx})^2}+\Big\{1+\frac{(1+\text{c}^2)}{(1+\text{cx})^2}\Big\}$
$=-\frac{(1+\text{x}^2)(1+\text{c}^2)}{(1+\text{cx})^2}+\frac{(1+\text{cx})^2+(\text{c}-\text{x})^2}{(1+\text{cx})^2}$
$=-\frac{(1+\text{x}^2)(1+\text{c}^2)}{(1+\text{cx})^2}+\frac{1+2\text{cx}+\text{c}^2\text{x}^2+\text{c}^2-2\text{cx}+\text{x}^2}{(1+\text{cx})^2}$
$=-\frac{(1+\text{x}^2)(1+\text{c}^2)}{(1+\text{cx})^2}+\frac{(1+\text{x})^2+\text{c}^2(1+\text{x})^2}{(1+\text{cx})^2}$
$=-\frac{(1+\text{x}^2)(1+\text{c}^2)}{(1+\text{cx})^2}+\frac{(1+\text{x}^2)(1+\text{c}^2)}{(1+\text{cx})^2}=0$
Hence, the given function is the solution to the given differential equation.
View full question & answer→Question 355 Marks
Find one-parameter families of solution curves of the following differential equation: (or solve the following differential equation)$\frac{\text{dy}}{\text{dx}}+3\text{y}=\text{e}^{\text{mx}},$ m is given real number.
AnswerWe have, $\frac{\text{dy}}{\text{dx}}+3\text{y}=\text{e}^{\text{mx}}\ \dots(1)$ Clearly, it is a linear differential equation of the form $\frac{\text{dy}}{\text{dx}}+\text{Py = Q}$ where $\text{P}=3$ $\text{Q}=\text{e}^{\text{mx}}$ $\therefore$ I.F. $=\text{e}^{\int\text{Pdx}}$ $=\text{e}^{\int3\text{dx}}$ $=\text{e}^{3\text{x}}$Multiplying both sides of (1) by $e^{3x}$, we get
$\text{e}^{3\text{x}}\Big(\frac{\text{dx}}{\text{dy}}+3\text{y}\Big)=\text{e}^{3\text{x}}\text{e}^{\text{mx}}$ $\Rightarrow\ \text{e}^{\text{3x}}\frac{\text{dx}}{\text{dy}}+3\text{e}^{3\text{x}}\text{y}=\text{e}^{(\text{m}+3)\text{x}}$ Integrating both sides with respect to x, we get $\text{ye}^{\text{3x}}=\int\text{e}^{(\text{m}+3)\text{x}}\text{dx + C}$ (when $\text{m}+3\neq0$) $\Rightarrow\ \text{ye}^{\text{3x}}=\frac{\text{e}^{(\text{m}+3)\text{x}}}{\text{m}+3}+\text{C}$ $\Rightarrow\ \text{y}=\frac{\text{e}^{\text{mx}}}{\text{m}+3}+\text{Ce}^{-3\text{x}}$ $\text{ye}^{3\text{x}}=\int\text{e}^{0\times\text{x}}\text{dx + C}$ (when $\text{m}+3=0$) $\Rightarrow\ \text{ye}^{3\text{x}}=\int\text{dx + C}$ $\Rightarrow\ \text{ye}^{3\text{x}}=\text{x + C}$ $\Rightarrow\ \text{y}=(\text{x + C})\text{e}^{-3\text{x}}$ Hence, $\text{y}=\frac{\text{e}^{\text{mx}}}{\text{m}+3}+\text{Ce}^{-3\text{x}},$ where $\text{m}+3\neq0$ and $\text{y}=(\text{x + C})\text{e}^{-3\text{x}},$ where $\text{m}+3=0$ are required solutions.
View full question & answer→Question 365 Marks
Show that the family of curves for which $\frac{\text{dy}}{\text{dx}}=\frac{\text{x}^2+\text{y}^2}{2\text{xy}},$ is given by $\text{x}^2-\text{y}^2=\text{Cx}$
AnswerThe given differential equation is
$\frac{\text{dy}}{\text{dx}}=\frac{\text{x}^2+\text{y}^2}{2\text{xy}}\ \dots(1)$
This is a homogeneous differential equation.
Putting y = vx and $\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}}$ in (1), we get
$\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{\text{x}^2+\text{v}^2\text{x}^2}{2\text{vx}^2}$
$\Rightarrow\ \text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{1+\text{v}^2}{2\text{v}}$
$\Rightarrow\ \frac{1+\text{v}^2}{2\text{v}}-\text{v}=\text{x}\frac{\text{dv}}{\text{dx}}$
$\Rightarrow\ \frac{1-\text{v}^2}{2\text{v}}=\text{x}\frac{\text{dv}}{\text{dx}}$
$\Rightarrow\ \frac{2\text{v}}{1-\text{v}^2}\text{dv}=\frac{\text{dx}}{\text{x}}$
Integrating on both sides, we get
$\int\frac{2\text{v}}{1-\text{v}^2}\text{dv}=\int\frac{\text{dx}}{\text{x}}$
$\Rightarrow\ \int\frac{-2\text{v}}{1-\text{v}^2}\text{dv}=-\int\frac{\text{dx}}{\text{x}}$
$\Rightarrow\ \log(1-\text{v}^2)=-\log\text{x}+\log\text{C}$
$\Rightarrow\ \log(1-\text{v}^2)+\log\text{x}=\log\text{C}$
$\Rightarrow\ \log(1-\text{v}^2)\text{x}=\log\text{C}$
$\Rightarrow\ (1-\text{v}^2)\text{x}=\text{C}$
$\Rightarrow\ \Big(1-\frac{\text{y}^2}{\text{x}^2}\Big)\text{x}=\text{C}$
$\Rightarrow\ \text{x}^2-\text{y}^2=\text{Cx}$
Thus, the family of curves for which $\frac{\text{dy}}{\text{dx}}=\frac{\text{x}^2+\text{y}^2}{2\text{xy}},$ is given by $\text{x}^2-\text{y}^2=\text{Cx}$
View full question & answer→Question 375 Marks
Solve the following differential equation:
$\text{x}\cos\Big(\frac{\text{y}}{\text{x}}\Big).(\text{y dx + x dy})=\text{y}\sin\Big(\frac{\text{y}}{\text{x}}\Big).(\text{x dy}-\text{y dx})$
AnswerWe have,
$\text{x}\cos\Big(\frac{\text{y}}{\text{x}}\Big).(\text{y dx + x dy})=\text{y}\sin\Big(\frac{\text{y}}{\text{x}}\Big).(\text{x dy}-\text{y dx})$
$\Rightarrow\ \text{xy}\cos\Big(\frac{\text{y}}{\text{x}}\Big)\text{dx}+\text{x}^2\cos\Big(\frac{\text{y}}{\text{x}}\Big)\text{dy}=\text{xy}\sin\Big(\frac{\text{y}}{\text{x}}\Big)\text{dy}-\text{y}^2\sin\Big(\frac{\text{y}}{\text{x}}\Big)\text{dx}$
$\Rightarrow\ \Big[\text{xy}\cos\Big(\frac{\text{y}}{\text{x}}\Big)+\text{y}^2\sin\Big(\frac{\text{y}}{\text{x}}\Big)\Big]\text{dx}=\Big[\text{xy}\sin\Big(\frac{\text{y}}{\text{x}}\Big)-\text{x}^2\cos\Big(\frac{\text{y}}{\text{x}}\Big)\Big]\text{dy}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{\text{xy}\cos\Big(\frac{\text{y}}{\text{x}}\Big)+\text{y}^2\sin\Big(\frac{\text{y}}{\text{x}}\Big)}{\text{xy}\sin\Big(\frac{\text{y}}{\text{x}}\Big)-\text{x}^2\cos\Big(\frac{\text{y}}{\text{x}}\Big)}$
This is a homogeneous differential equation.
Putting x = vy and $\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}}$, we get
$\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{\text{vx}^2\cos\text{v}+\text{v}^2\text{x}^2\sin\text{v}}{\text{vx}^2\sin\text{v}-\text{x}^2\cos\text{v}}$
$\Rightarrow\ \text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{\text{v}\cos\text{v}+\text{v}^2\sin\text{v}}{\text{v}\sin\text{v}-\cos\text{v}}$
$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{\text{v}\cos\text{v}+\text{v}^2\sin\text{v}}{\text{v}\sin\text{v}-\cos\text{v}}-\text{v}$
$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{\text{v}\cos\text{v}+\text{v}^2\sin\text{v}-\text{v}^2\sin\text{v}+\text{v}\cos\text{v}}{\text{v}\sin\text{v}-\cos\text{v}}$
$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{2\text{v}\cos\text{v}}{\text{v}\sin\text{v}-\cos\text{v}}$
$\Rightarrow\ \frac{\text{v}\sin\text{v}-\cos\text{v}}{2\text{v}\cos\text{v}}\text{dv}=\frac{1}{\text{x}}\text{dx}$
Integrating both sides, we get
$\int\frac{\text{v}\sin\text{v}-\cos\text{v}}{2\text{v}\cos\text{v}}\text{dv}=\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\ \int\frac{\text{v}\sin\text{v}-\cos\text{v}}{\text{v}\cos\text{v}}\text{dv}=2\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\ \int\frac{\text{v}\sin\text{v}}{\text{v}\cos\text{v}}\text{dv}-\int\frac{\cos\text{v}}{\text{v}\cos\text{v}}\text{dv}=2\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\ \int\tan\text{v dv}-\int\frac{1}{\text{v}}\text{dv}=2\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\ \log|\sec\text{v}|-\log|\text{v}|=2\log|\text{x}|+\log\text{C}$
$\Rightarrow\ \log\Big|\frac{\sec\text{v}}{\text{v}}\Big|=\log\big|\text{Cx}^2\big|$
$\Rightarrow\ \frac{\sec{\text{v}}}{\text{v}}=\text{Cx}^2$
Putting $\text{v}=\frac{\text{y}}{\text{x}}$, we get
$\sec\Big(\frac{\text{y}}{\text{x}}\Big)=\frac{\text{y}}{\text{x}}\times\text{C}\times\text{x}^2$
$\Rightarrow\ \sec\Big(\frac{\text{y}}{\text{x}}\Big)=\text{Cxy}$
Hence, $\sec\Big(\frac{\text{y}}{\text{x}}\Big)=\text{Cxy}$ is the required solution.
View full question & answer→Question 385 Marks
Form the differential equation of the family of hyperebolas having foci on x- axis and centre at the origine.
AnswerThe equation of the family of hyperbolas having the centre at the origin and foci on the x-axis is
$\frac{\text{x}^2}{\text{a}^2}-\frac{\text{y}^2}{\text{b}^2}=1\ ...(1)$
where a and b are parameters. As this equation contains two parameters, we shall get a second-order differential equation.
Differentiating equation (1) with respect to x, we get
$\frac{2\text{x}}{\text{a}^2}-\frac{2\text{y}}{\text{b}^2}\frac{\text{dy}}{\text{dx}}=0\ ...(2)$
Differentiating equation (2) with respect to x, we get
$\frac{2}{\text{a}^2}-\frac{2}{\text{b}^2}\Big[\text{y}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2\Big]=0$
$\Rightarrow\frac{1}{\text{a}^2}=\frac{1}{\text{b}^2}\Big[\text{y}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2\Big]$
$\Rightarrow\frac{\text{b}^2}{\text{a}^2}=\Big[\text{y}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2\Big]$
Now, from equation (2), we get
$\frac{2\text{x}}{\text{a}^2}=\frac{2\text{y}}{\text{b}^2}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\frac{\text{b}^2}{\text{a}^2}=\frac{\text{y}}{\text{x}}\frac{\text{dy}}{\text{dx}}\ ...(4)$
From (3) and (4), we get
$\frac{\text{y}}{\text{x}}\frac{\text{dy}}{\text{dx}}=\Big[\text{y}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2\Big]$
$\Rightarrow\text{y}\frac{\text{dy}}{\text{dx}}=\text{xy}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{x}\Big(\frac{\text{dy}}{\text{dx}}\Big)^2$
$\Rightarrow\text{xy}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{x}\Big(\frac{\text{dy}}{\text{dx}}\Big)^2-\text{y}\frac{\text{dy}}{\text{dx}}=0$
It is the required differential equation.
View full question & answer→Question 395 Marks
Solve the following differential equation
$\text{e}^{\frac{\text{dy}}{\text{dx}}}=\text{x}+1;\text{y}(0)=3$
AnswerWe have
$\text{e}^{\frac{\text{dy}}{\text{dx}}}=\text{x}+1$
Taking log on both sides, we get
$\frac{\text{dy}}{\text{dx}}\log\text{e}=\log(\text{x}+1)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\log(\text{x}+1) $
$\Rightarrow\text{dy}=\{\log(\text{x}+1)\}\text{dx}$
intregrating both sides ,we get
$\int\text{dy}=\int\{\log(\text{x}+1)\}\text{dx}$
$\Rightarrow\text{y}=\int1\times\log(\text{x}+1)\text{dx}$
$\Rightarrow\text{y}=\log(\text{x}+1)\int1\text{dx}-\int\Big[\frac{\text{d}}{\text{dx}}(\log\text{x}+1)\int1\text{dx}\Big]\text{dx}$
$\Rightarrow\text{y}=\text{x}\log(\text{x}+1)-\int\frac{\text{x}}{\text{x}+1}\ \text{dx}$
$\Rightarrow\text{y}=\text{x}\log(\text{x}+1)-\int\Big(1-\frac{1}{\text{x}+1}\Big)\text{dx}$
$\Rightarrow\text{y}=\text{x}\log(\text{x}+1)-\text{x}+\log(\text{x}+1)+\text{C}\ ...(1)$
It is given that y(0) = 3
$\therefore3=0\times\log(0+1)-0+\log(0+1)+\text{C}$
$\Rightarrow\text{C}=3$
Substituting the value of C in (1), we get
$\text{y}=\text{x}\log(\text{x}+1)+\log(\text{x}+1)-\text{x}+3$
$\Rightarrow\text{y}=(\text{x}+1)\log(\text{x}+1)-\text{x}+3$
Hence, $\text{y}=(\text{x}+1)\log(\text{x}+1)-\text{x}+3$is the solution to the given differential equation.
View full question & answer→Question 405 Marks
Find one-parameter families of solution curves of the following differential equation: (or solve the following differential equation)$\text{x}\frac{\text{dy}}{\text{dx}}+2\text{y}=\text{x}^2\log\text{x}$
AnswerWe have$\text{x}\frac{\text{dy}}{\text{dx}}+2\text{y}=\text{x}^2\log\text{x}$
Dividing both sides by x, we get$\frac{\text{dy}}{\text{dx}}+\frac{2\text{y}}{\text{x}}=2\log\text{x}$
Comparing with $\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q},$ we get
$\text{P}=\frac{2}{\text{x}}$
$\text{Q}=\text{x}\log\text{x}$
Now,$\text{I.F}=\text{e}^{\int\text{Pdx}}$
$=\text{e}^{\int\frac{2\text{}}{\text{x}}}\text{ dx}$
$=\text{e}^{2\log|\text{x}|}$
$=\text{x}^2$
So, the solution is given by$\text{y}\times\text{I.F.}=\int\text{Q}\times\text{I.F.}\text{ dx}+\text{C}$
$\Rightarrow\text{x}^2\text{y}=\int\text{x}^3\log\text{x dx}+\text{C}$
$\Rightarrow\text{x}^2\text{y}=\log\text{x}\int\text{x}^3\text{dx}-\int\Big[\frac{\text{d}}{\text{dx}}(\log\text{x})\int\text{x}^3\text{dx}\Big]\text{dx}+\text{C}$
$\Rightarrow\text{x}^2\text{y}=\frac{\text{x}^4\log\text{x}}{4}-\int\frac{\text{x}^3}{4}\text{dx}+\text{C}$
$\Rightarrow\text{x}^2\text{y}=\frac{\text{x}^4\log\text{x}}{4}-\frac{\text{x}^3}{16}\text{dx}+\text{C}$
$\Rightarrow\text{y}=\frac{\text{x}^2\log\text{x}}{4}-\frac{\text{x}^2}{16}+\frac{\text{C}}{\text{x}^2}$
$\Rightarrow\text{y}=\frac{\text{x}^2}{16}(4\log\text{x}-1)+\frac{\text{C}}{\text{x}^2}$
View full question & answer→Question 415 Marks
Solve the following initial value problems:
$\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}=\text{x}\cos\text{x}+\sin\text{x},\text{ y}\Big(\frac{\pi}{2}\Big)=1$
AnswerWe have,
$\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}=\text{x}\cos\text{x}+\sin\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}+\frac{1}{\text{x}}\text{y}=\cos\text{x}+\frac{\sin\text{x}}{\text{x}}\ ...(1)$
Clearly, it is a linear differential equation of the form
$\frac{\text{dy}}{\text{dx}}+\text{Px}=\text{Q}$
Where $\text{P}=\tan\text{x}$ and $\text{Q}=\text{x}^2\cot\text{x}+2\text{x}$
$\therefore\text{ I.F.}=\text{e}^{\int\text{Pdx}}$
$=\text{e}^{\int\frac{1}{\text{x}}\text{dx}}$
$=\text{e}^{\log\text{x}}$
$=\text{x}$
Multiplying both sides of (1) by $\text{I.F.}=\text{x},$ we get
$\text{x}\Big(\frac{\text{dy}}{\text{dx}}+\frac{1}{\text{x}}\text{y}\Big)=\text{x}\Big(\cos\text{x}+\frac{\sin\text{x}}{\text{x}}\Big)$
$\Rightarrow\text{x}\Big(\frac{\text{dy}}{\text{dx}}+\frac{1}{\text{x}}\text{y}\Big)=\text{x}\cos\text{x}+\sin\text{x}$
Integrating both sides with respect to x, we get
$\text{xy}=\int\text{x}\cos\text{x dx}+\int\sin\text{x dx}+\text{C}$
$\Rightarrow\text{xy}=\Big[\text{x}\sin\text{x}-\int1(\sin\text{x})\text{dx}\Big]-\cos\text{x}+\text{C}$
$\Rightarrow\text{xy}=\text{x}\sin\text{x}+\cos\text{x}-\cos\text{x}+\text{C}$
$\Rightarrow\text{xy}=\text{x}\sin\text{x}+\text{C}\ ...(2)$
Now,
$\text{y}\Big(\frac{\pi}{2}\Big)=1$
$\therefore\ 1\times\frac{\pi}{2}=\frac{\pi}{2}\sin\frac{\pi}{2}+\text{C}$
$\Rightarrow\text{C}=0$
Putting the value of C in (2), we get
$\text{xy}=\text{x}\sin\text{x}$
$\Rightarrow\text{y}=\sin\text{x}$
Hence, $\text{y}=\sin\text{x}$ is the required solution.
View full question & answer→Question 425 Marks
Find one-parameter families of solution curves of the following differential equation: (or solve the following differential equation)$\frac{\text{dy}}{\text{dx}}-\frac{2\text{xy}}{1+\text{x}^2}=\text{x}^2+2$
AnswerWe have,
$\frac{\text{dy}}{\text{dx}}-\frac{2\text{xy}}{1+\text{x}^2}=\text{x}^2+2\ \dots(1)$
Clearly, it is a linear differential equation of the form
$\frac{\text{dy}}{\text{dx}}+\text{Py = Q}$
where
$\text{P}=-\frac{2\text{x}}{1+\text{x}^2}$
$\text{Q}=\text{x}^2+2$
$\therefore$ I.F. $=\text{e}^{\int\text{Pdx}}$
$=\text{e}^{-\int\frac{2\text{x}}{1+\text{x}^2}\text{dx}}$
$=\text{e}^{-\log|1+\text{x}^2|}$
$=\frac{1}{1+\text{x}^2}$
Multiplying both sides of (1) by $\frac{1}{1+\text{x}^2},$ we get
$\frac{1}{1+\text{x}^2}\Big(\frac{\text{dy}}{\text{dx}}-\frac{2\text{xy}}{1+\text{x}^2}\Big)=\frac{1}{1+\text{x}^2}(\text{x}^2+2)$
$\Rightarrow\ \frac{1}{1+\text{x}^2}\frac{\text{dy}}{\text{dx}}-\frac{2\text{xy}}{(1+\text{x}^2)^2}=\frac{\text{x}^2+2}{\text{x}^2+1}$
Integrating both sides with respect to x, we get
$\frac{1}{1+\text{x}^2}\text{y}=\int\frac{\text{x}^2+2}{\text{x}^2+1}\text{dx + C}$
$\Rightarrow\ \frac{1}{1+\text{x}^2}\text{y}=\int\frac{\text{x}^2+1+1}{\text{x}^2+1}\text{dx + C}$
$\Rightarrow\ \frac{1}{1+\text{x}^2}\text{y}=\int\text{dx}+\int\frac{1}{\text{x}^2+1}\text{dx + C}$
$\Rightarrow\ \frac{1}{1+\text{x}^2}\text{y}=\text{x}+\tan^{-1}\text{x}+\text{C}$
$\Rightarrow\ \text{y}=(1+\text{x}^2)(\text{x}+\tan^{-1}+\text{C})$
Hence, $\text{y}=(1+\text{x}^2)(\text{x}+\tan^{-1}+\text{C})$ is the required solution.
View full question & answer→Question 435 Marks
Solve the following initial value problems:
$\text{x}(\text{x}^2+3\text{y}^2)\text{dx}+\text{y}(\text{y}^2+3\text{x}^2)\text{dy}=0,\text{y}(1)=1$
Answer$\text{x}(\text{x}^2+3\text{y}^2)\text{dx}+\text{y}(\text{y}^2+3\text{x}^2)\text{dy}=0,\text{y}(1)=1$
$\frac{\text{dy}}{\text{dx}}=-\frac{\text{x}(\text{x}^2+3\text{y}^2)}{\text{y}(\text{y}^2+3\text{x}^2)}$
It is a homogeneous equation
put y = vx
$\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}}$
So,
$\text{v + x}\frac{\text{dv}}{\text{dx}}=-\frac{\text{x}(\text{x}^2+3\text{v}^2\text{x}^2)}{\text{vx}(\text{v}^2\text{x}^2+3\text{x}^2)}$
$\text{x}\frac{\text{dv}}{\text{dx}}=\frac{(1+3\text{v}^2)}{\text{v}(\text{v}^2+3)}-\text{v}$
$\text{x}\frac{\text{dv}}{\text{dx}}=\frac{-1-3\text{v}^2-\text{v}^4-3\text{v}^2}{\text{v}(\text{v}^2+3)}$
$=\frac{-\text{v}^4-6\text{v}^2-1}{\text{v}(\text{v}^2+3)}$
$\frac{\text{v}(\text{v}^2+3)}{\text{v}^4+6\text{v}^2+1}\text{dv}=-\frac{\text{dx}}{\text{x}}$
$\int\frac{4\text{v}^3+12\text{v}}{\text{v}^4+6\text{v}^2+1}\text{dv}=-4\int\frac{\text{dx}}{\text{x}}$
$\log|\text{v}^4+6\text{v}^2+1|=\log\Big|\frac{\text{C}}{\text{x}^4}\Big|$
$|\text{v}^4+6\text{v}^2+1|=\Big|\frac{\text{C}}{\text{x}^4}\Big|\ \dots(\text{i})$
Put y = 1, x = 1
(1+6+1) = C
⇒ C = 8
Put C = 8 in equation (i),
$(\text{y}^4+\text{x}^4+6\text{x}^2\text{y}^2)=8$
View full question & answer→Question 445 Marks
Solve the following initial value problems:
$\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}=(\text{x}+1)\text{e}^{\text{x}},\text{ y}(1)=0$
AnswerWe have,
$\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}=(\text{x}+1)\text{e}^{\text{x}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}-\frac{1}{\text{x}}\text{y}=\Big(\frac{\text{x}+1}{\text{x}}\Big)\text{e}^{\text{x}}\ ....(1)$
Clearly, it is a linear differential equation of the form
$\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
Where $\text{P}=-\frac{1}{\text{x}}$ and $\text{Q}=\frac{\text{x}+1}{\text{x}}\text{e}^{-\text{x}}$
$\therefore\text{ I.F.}=\text{e}^{\int\text{Pdx}}$
$=\text{e}^{-\int\frac{1}{\text{x}}\text{dx}}$
$=\text{e}^{\log\text{x}}$
$=\frac{1}{\text{x}}$
Multiplying both sides of (1) by $\text{I.F.}=\frac{1}{\text{x}},$ we get
$\frac{1}{\text{x}}\Big(\frac{\text{dy}}{\text{dx}}-\frac{1}{\text{x}}\text{y}\Big)=\Big(\frac{\text{x}+1}{\text{x}^2}\Big)\text{e}^{-\text{x}}$
Integrating both sides with respect to x, we get
$\frac{1}{\text{x}}\text{y}=\int\Big(\frac{1}{\text{x}}+\frac{1}{\text{x}^2}\Big)\text{e}^{-\text{x}}\text{dx}+\text{C}$
Putting $\frac{1}{\text{x}}\text{e}^{-\text{x}}=\text{t}$
$\Rightarrow\Big(-\frac{1}{\text{x}}\text{e}^{-\text{x}}-\frac{1}{\text{x}^2}\text{e}^{-\text{x}}\Big)\text{dx}=\text{dt}$
$\Rightarrow\Big(\frac{1}{\text{x}}+\frac{1}{\text{x}^2}\Big)\text{e}^{-\text{x}}\text{dx}=-\text{dt}$
$\therefore\ \frac{1}{\text{x}}\text{y}=\int-\text{dt}+\text{C}$
$\Rightarrow\frac{\text{y}}{\text{x}}=-\text{t}+\text{C}$
$\Rightarrow\frac{\text{y}}{\text{x}}=-\frac{\text{e}^{\text{x}}}{\text{x}}+\text{C}$
$\Rightarrow\text{y}=-\text{e}^{-\text{x}}+\text{Cx}\ ...(2)$
Now,
$\text{y}(1)=0$
$\therefore0=-\text{e}^{-1}+\text{C}$
$\Rightarrow\text{y}=\text{xe}^{-1}-\text{e}^{-\text{x}}$
Hence, $\text{y}=\text{xe}^{-1}-\text{e}^{-\text{x}}$ is the required solution.
View full question & answer→Question 455 Marks
Solve the following initial value problems:
$\frac{\text{dy}}{\text{dx}}+\text{y}\cot\text{x}=4\text{x }\text{cosec x},\text{ y}\Big(\frac{\pi}{2}\Big)=0$
AnswerWe have,
$\frac{\text{dy}}{\text{dx}}+\text{y}\cot\text{x}=4\text{x }\text{cosec x}\ ...(1)$
Clearly, it is a linear differential equation of the form
$\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
Where $\text{P}=\cot\text{x}$ and $\text{Q}=4\text{x cosec x}$
$\therefore\text{ I.F.}=\text{e}^{\int\text{Pdx}}$
$=\text{e}^{\int\cos\text{x dx}}$
$=\text{e}^{\log|\sin\text{x}|}$
$=\sin\text{x}$
Multiplying both sides of (1) by $\text{I.F.}=\sin\text{x},$ we get
$\sin\text{x}\Big(\frac{\text{dy}}{\text{dx}}+\text{y}\cot\text{x}\Big)=\sin\text{x}(4\text{x cosec x})$
$\Rightarrow\sin\text{x}\Big(\frac{\text{dy}}{\text{dx}}+\text{y}\cot\text{x}\Big)=4\text{x}$
Integrating both sides with respect to x, we get
$\text{y}\sin\text{x}=4\int\text{x dx}+\text{C}$
$\Rightarrow\text{y}\sin\text{x}=2\text{x}^2+\text{C}\ ...(2)$
Now,
$\text{y}\Big(\frac{\pi}{2}\Big)=0$
$\therefore\ 0\times\sin\Big(\frac{\pi}{2}\Big)=2\Big(\frac{\pi}{2}\Big)^2+\text{C}$
$\Rightarrow\text{C}=-\frac{\pi^2}{2}$
Putting the value of C in (2) we get
$\text{y}\sin\text{x}=2\text{x}^2-\frac{\pi^2}{2}$
Hence, $\text{y}\sin\text{x}=2\text{x}^2-\frac{\pi^2}{2}$ is the required solution.
View full question & answer→Question 465 Marks
Solve the following differential equation:
$2\text{xy dx}+(\text{x}^2+2\text{y}^2)\text{dy}=0$
AnswerHere, $2\text{xy dx}+(\text{x}^2+2\text{y}^2)\text{dy}=0$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{2\text{xy}}{\text{x}^2-2\text{y}^2}$
It is a homogeneous equation.
Put y = vx and $\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}}$
So,
$\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{2\text{xvx}}{\text{x}^2+2\text{v}^2\text{x}^2}$
$\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{2\text{v}}{1+2\text{v}^2}$
$\text{x}\frac{\text{dv}}{\text{dx}}=\frac{2\text{v}}{1-2\text{v}^2}-\text{v}$
$=\frac{2\text{v}-\text{v}+2\text{v}^3}{1+2\text{v}^2}$
$\text{x}\frac{\text{dv}}{\text{dx}}=\frac{\text{v}-2\text{v}^3}{1+2\text{v}^2}$
$\int\frac{1+2\text{v}^2}{\text{v}-2\text{v}^3}\text{dv}=\int\frac{\text{dx}}{\text{x}}\ \dots(\text{i})$
$\frac{1+2\text{v}^2}{\text{v}-2\text{v}^3}=\frac{1+2\text{v}^2}{\text{v}(1-2\text{v}^2)}$
$\frac{1+2\text{v}^2}{\text{v}(1-2\text{v}^2)}=\frac{\text{A}}{\text{v}}+\frac{\text{Bv + C}}{1-2\text{v}^2}$
$\frac{1+2\text{v}^2}{\text{v}(1-2\text{v}^2)}=\frac{\text{A}(1-2\text{v}^2)+(\text{Bv + C)}\text{v}}{\text{v}(1-2\text{v}^2)}$
$1+2\text{v}^2=\text{A}-2\text{Av}^2+\text{Bv}^2+\text{Cv}$
$1+2\text{v}^2=\text{v}^2(-2\text{A + B})+\text{Cv + A}$
Comparing the co-efficients of like powers of v,
A = 1
C = 0
-2A + B = 2
-2 + B = 0
B = 4
$\frac{1+2\text{v}^2}{\text{v}-2\text{v}^3}=\frac{1}{\text{v}}+\frac{4\text{v}}{1-2\text{v}^2}$
$\frac{1+2\text{v}^2}{\text{v}-2\text{v}^3}=\frac{1}{\text{v}}-\frac{(-4\text{v})}{(1-2\text{v}^2)}$
View full question & answer→Question 475 Marks
Find the particular solution of the differential equation $\text{x}\cos\Big(\frac{\text{y}}{\text{x}}\Big)\frac{\text{dy}}{\text{dx}}=\text{y}\cos\Big(\frac{\text{y}}{\text{x}}\Big)+\text{x},$ given that when $\text{x}=1,\text{y}=\frac{\pi}4$.
Answer$\text{x}\cos\Big(\frac{\text{y}}{\text{x}}\Big)\frac{\text{dy}}{\text{dx}}=\text{y}\cos\Big(\frac{\text{y}}{\text{x}}\Big)+\text{x}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{\text{y}\cos\Big(\frac{\text{y}}{\text{x}}\Big)+\text{x}}{\text{x}\cos\Big(\frac{\text{y}}{\text{x}}\Big)}$
This is a homogeneous differential equation.
puttuing y = vx and $\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}},$ we get
$\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{\text{vx}+\cos\text{v + x}}{\text{x}\cos\text{v}}$
$\Rightarrow\ \text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{\text{v}+\cos\text{v + 1}}{\cos\text{v}}$
$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{\text{v}\cos\text{v}+1-\text{v}\cos\text{v}}{\cos{\text{v}}}$
$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{1}{\cos\text{v}}$
$\Rightarrow\ \cos\text{v dv}=\frac{1}{\text{x}}\text{dx}$
Integrating both sides, we get
$\int\cos\text{v dv}=\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\ \sin\text{v}=\log|\text{x}|+\text{C}$
Putting $\text{v}=\frac{\text{y}}{\text{x}},$ we get
$\sin\frac{\text{y}}{\text{x}}=\log|\text{x}|+\text{C}\ \dots(1)$
At $\text{x}=1,\text{y}=\frac{\pi}4$ (Given)
Putting $\text{x}=1$ and $\text{y}=\frac{\pi}4$ in (1), we get
$\text{C}=\frac{1}{\sqrt2}$
Putting $\text{C}=\frac{1}{\sqrt2}$ in (1), we get
$\sin\frac{\text{y}}{\text{x}}=\log|\text{x}|+\frac{1}{\sqrt2}$
Hence, $\sin\frac{\text{y}}{\text{x}}=\log|\text{x}|+\frac{1}{\sqrt2}$ is the required solution.
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Solve the following differential equation:
$(1+\text{y}^2)+(\text{x}-\text{e}^{\tan^{-1}\text{y}})\frac{\text{dy}}{\text{dx}}=0$
AnswerWe have, $(1+\text{y}^2)+(\text{x}-\text{e}^{\tan^{-1}\text{y}})\frac{\text{dy}}{\text{dx}}=0$ $\Rightarrow\ (\text{x}-\text{e}^{\tan^{-1}\text{y}})\frac{\text{dy}}{\text{dx}}=-(1+\text{y}^2)$ $\Rightarrow\ \frac{\text{dy}}{\text{dx}}=-\frac{(1+\text{y}^2)}{(\text{x}-\text{e}^{\tan^{-1}\text{y}})}$$\Rightarrow\ \frac{\text{dx}}{\text{dy}}=-\frac{\text{x}-\text{e}^{\tan^{-1}\text{y}}}{1+\text{y}^2}$
$\Rightarrow\ \frac{\text{dx}}{\text{dy}}+\frac{\text{x}}{1+\text{y}^2}=\frac{\text{e}^{\tan^{-1}\text{y}}}{1+\text{y}^2}\ \dots(1)$
Clearly, it is a linear differential equation of the form $\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$ where $\text{P}=\frac{1}{1+\text{y}^2}$ $\text{Q}=\frac{\text{e}^{\tan^{-1}\text{y}}}{1+\text{y}^2}$ $\therefore$ I.F. $=\text{e}^{\int\text{Pdy}}$ $=\text{e}^{\int\frac{1}{1+\text{y}^2}\text{dy}}$ $=\text{e}^{{\tan^{-1}\text{y}}}$ Multiplying both sides of (1) by $\text{e}^{\tan^{-1}\text{y}},$ we get $\text{e}^{\tan^{-1}\text{y}}\Big(\frac{\text{dx}}{\text{dy}}+\frac{\text{x}}{1+\text{y}^2}\Big)=\text{e}^{\tan^{-1}\text{y}}\frac{\text{e}^{\tan^{-1}\text{y}}}{1+\text{y}^2}$ $\Rightarrow\ \text{e}^{\tan^{-1}\text{y}}\frac{\text{dx}}{\text{dy}}+\frac{\text{x}\text{e}^{\tan^{-1}\text{x}}}{1+\text{y}^2}=\frac{\text{e}^{\tan^{-1}\text{y}}}{1+\text{y}^2}$ Integrating both sides with respect to y, we get $\text{x}\text{e}^{\tan^{-1}\text{y}}=\int\frac{\text{e}^{2\tan^{-1}\text{y}}}{1+\text{y}^2}\text{dy + C}$ $\Rightarrow\ \text{x}\text{e}^{\tan^{-1}\text{y}}=\text{I + C}\ \dots(2)$ Here, $\text{I}=\int\frac{\text{e}^{2\tan^{-1}\text{y}}}{1+\text{y}^2}\text{dy}$ Putting $\tan^{-1}\text{y = t},$ we get $\frac{1}{1+\text{y}^2}\text{dy = dt}$ $\therefore\ \text{I}=\int\text{e}^{2\text{t}}\text{dt}$ $=\frac{\text{e}^{2\text{t}}}{2}$ $=\frac{\text{e}^{2\tan^{-1}\text{y}}}{2}$ Putting the value of I in (2), we get $\text{x}\text{e}^{\tan^{-1}\text{y}}=\frac{\text{e}^{2\tan^{-1}\text{y}}}{2}+\text{C}$ $\Rightarrow\ 2\text{x}\text{e}^{\tan^{-1}\text{y}}=\text{e}^{2\tan^{-1}\text{y}}+2\text{C}$ $\Rightarrow\ 2\text{x}\text{e}^{\tan^{-1}\text{y}}=\text{e}^{2\tan^{-1}\text{y}}+\text{K}$ (where K = 2C) Hence, $2\text{x}\text{e}^{\tan^{-1}\text{y}}=\text{e}^{2\tan^{-1}\text{y}}+\text{K}$ is the required solution.
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Solve the following differential equation
$\frac{\text{dy}}{\text{dx}}=\cos^3\text{x}\sin^2\text{x}+\text{x}\sqrt{2\text{x}+1}$
AnswerWe have,
$\frac{\text{dy}}{\text{dx}}=\cos^3\text{x}\sin^2\text{x}+\text{x}\sqrt{2\text{x}+1}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=(\cos^3\text{x}\sin^2\text{x}+\text{x}\sqrt{2\text{x}+1})\text{dx}$
Integrating both sides, we get
$\int\frac{\text{dy}}{\text{dx}}=\int(\cos^3\text{x}\sin^2\text{x}+\text{x}\sqrt{2\text{x}+1})\text{dx}$
$\Rightarrow\text{y}=\int\cos^3\text{x}\sin^2\text{x dx}+\int\text{x}\sqrt{2\text{x}+1}\text{dx}$
$\Rightarrow\text{y}=\text{I}_1+\text{I}_2\ ...(1)$
Where
$\text{I}_1=\int\cos^3\text{x}\sin^2\text{x dx}$
$\text{I}_2=\int\text{x}\sqrt{2\text{x}+1}\text{dx}$
Now,
$\text{I}_1=\int\cos^3\text{x}\sin^2\text{x dx}$
$=\int\sin^2\text{x}(1-\sin^2\text{x})\cos\text{x dx}$
Putting $\text{t}=\sin\text{x},$ we get
$\text{dt}=\cos\text{x dx}$
$\Rightarrow\text{I}_1=\int\text{t}^2(1-\text{t}^2)\text{dt}$
$=\int(\text{t}^2-\text{t}^4)\text{dt}$
$=\frac{\text{t}^3}{3}-\frac{\text{t}^5}{5}+\text{C}_1$
$=\frac{\sin^3\text{x}}{3}-\frac{\sin^5\text{x}}{5}+\text{C}_1$
$\text{I}_2=\int\text{x}\sqrt{2\text{x}+1}\text{dx}$
Putting $\text{t}^2=2\text{x}+1$ we get,
$2\text{t dt}=2\text{dx}$
$\Rightarrow\text{t dt}=\text{dx}$
Now,
$\text{I}_2=\int\Big(\frac{\text{t}^2-1}{2}\Big)\text{t}\times\text{t}\text{ dt}$
$=\frac{1}{2}\int(\text{t}^4-\text{t}^2)\text{dt}$
$=\frac{\text{t}^5}{10}-\frac{\text{t}^3}{6}+\text{C}_2$
$=\frac{(2\text{x}+1)\frac{5}{2}}{10}-\frac{(2\text{x}+1)^{\frac{3}{2}}}{6}+\text{C}_2$
Putting the value of $I_1$ and $I_2$ in (1), we get
$\text{y}=\frac{\sin^3\text{x}}{3}-\frac{\sin^5\text{x}}{5}+\text{C}_1+\frac{(2\text{x}+1)^{\frac{5}{2}}}{6}+\text{C}_2$
$\text{y}=\frac{\sin^3\text{x}}{3}-\frac{\sin^5\text{x}}{5}+\frac{(2\text{x}+1)^{\frac{5}{2}}}{10}-\frac{(2\text{x}+1)^{\frac{3}{2}}}{6}+\text{C}$
Hence, $\text{y}=\frac{\sin^3\text{x}}{3}-\frac{\sin^5\text{x}}{5}+\frac{(2\text{x}+1)^{\frac{5}{2}}}{10}-\frac{(2\text{x}+1)^{\frac{3}{2}}}{6}+\text{C}$ is the solution to the given differential equation.
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Show that $\text{y}=4\text{ax}$ is a solution of the differential equation $\text{y}=\text{x}\frac{\text{d}\text{y}}{\text{dx}}+\text{a}\frac{\text{dx}}{\text{dy}}.$
AnswerWe have,
$\text{y}=4\text{ax}\ ...(1)$
Differentiating both sides of equation (1) with respect to 3, we get
$2\text{y}\frac{\text{dy}}{\text{dx}}=4\text{a}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{2\text{a}}{\text{y}}\ ...(2)$
now, differentiating both sided of (1) with respect to y, we get
$2\text{y}=4\text{a}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{2\text{a}}\ ...(3)$
$\therefore\text{x}\frac{\text{dy}}{\text{dx}}+\text{a}\frac{\text{dy}}{\text{dx}}=\text{x}\Big(\frac{2\text{a}}{\text{y}}\Big)+\text{a}\Big(\frac{\text{y}}{2\text{a}}\Big)$
$\Rightarrow\text{x}\frac{\text{dy}}{\text{dx}}+\text{a}\frac{\text{dy}}{\text{dx}}=\frac{2\text{ax}}{\text{y}}+\frac{\text{y}}{2}$
$\Rightarrow\text{x}\frac{\text{dy}}{\text{dx}}+\text{a}\frac{\text{dy}}{\text{dx}}=\frac{\text{y}^2}{2\text{y}}+\frac{\text{y}}{2}$
$\Rightarrow\text{x}\frac{\text{dy}}{\text{dx}}+\text{a}\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{2}+\frac{\text{y}}{2}$
$\Rightarrow\text{x}\frac{\text{dy}}{\text{dx}}+\text{a}\frac{\text{dy}}{\text{dx}}=\text{y}$
$\Rightarrow\text{y}=\text{x}\frac{\text{dy}}{\text{dx}}+\text{a}\frac{\text{dy}}{\text{dx}}$
Hence, the given function is the solution to the given differential equation.
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