Question 15 Marks
Form the differential equation representing the family of ellipses having centre at the origin and foci on x-axis.
Answerwe know that the equation of said family of ellopsis is
$\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=1\ ...(1)$
Differentiating (1) w..r.t.x, we get
$\frac{2\text{x}}{\text{a}^2}+\frac{2\text{y}}{\text{b}^2}.\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\frac{\text{y}}{\text{x}}\Big(\frac{\text{dy}}{\text{dx}}\Big)=\frac{-\text{b}^2}{\text{a}^2}\ ...(2)$
Differentiating (2) w..r.t.x, we get
$\frac{\text{y}}{\text{x}}\Big(\frac{\text{d}^2\text{y}}{\text{dx}^2}\Big)+\bigg(\frac{\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}}{\text{x}^2}\bigg)\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\text{xy}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{x}\Big(\frac{\text{dy}}{\text{dx}}\Big)^2-\text{y}\frac{\text{dy}}{\text{dx}}$
Which is the required difeerential equation.
View full question & answer→Question 25 Marks
Find the equation of the curve which passes through the point (1, 2) and the distance between the foot of the ordinate of the point of contact and the point of intersection of the tangent with x-axis twice abscissa of the pont of contact.
Answer
It is given that the distance between the foot of ordinate of point of contanct (A) and the point of intersection of tangent with x-axis (T) = 2x
Coordinate of $\text{T}=\big(\text{x}-\text{y}\frac{\text{dx}}{\text{dy}}, 0\big)$
$\text{AT}=\Big[\text{x}-\Big(\text{x}-\text{y}\frac{\text{dx}}{\text{dy}}, 0\Big)\Big]=2\text{x}$
Equation of tangent,
$\text{y}-\text{y}=\frac{\text{dy}}{\text{dx}}(\text{x}-\text{x})$
$\Rightarrow\text{y}-\text{0}=\frac{\text{dy}}{\text{dx}}\Big(\text{x}-\big(\text{x}-\text{y}\frac{\text{dx}}{\text{dy}}\big)\Big)$
$\Rightarrow \text{y}\frac{\text{dx}}{\text{dy}}=2\text{x}$
$\Rightarrow \int\frac{\text{dx}}{\text{x}}=2\int\frac{\text{dy}}{\text{y}}$
$\Rightarrow \log\text{x}=\log\text{y}^{2}+\log\text{C}$
$\text{x}=\text{Cy}^{2}$
As the circle passes through (1, 2)
$1=\text{C}\times2^{2}$
$\Rightarrow \text{C}=\frac{1}{4}$
$\Rightarrow 4\text{x}=\text{y}^{2}$ View full question & answer→Question 35 Marks
Solve the following differential equations:
$\frac{\text{dy}}{\text{dx}}=\frac{\text{y}^2-\text{x}^2}{2\text{xy}}$
AnswerHere, $\frac{\text{dy}}{\text{dx}}=\frac{\text{y}^2-\text{x}^2}{2\text{xy}}$It is homogeneous equation
Put y = vx
$\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}}$
So,
$\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{\text{v}^2\text{x}^2-\text{x}^2}{2\text{xvx}}$
$\text{x}\frac{\text{dv}}{\text{dx}}=\frac{\text{v}^2-1}{2\text{v}}-\frac{\text{v}}1$
$=\frac{\text{v}^2-1-2\text{v}^2}{2\text{v}}$
$\text{x}\frac{\text{dv}}{\text{dx}}=\frac{-1-\text{v}^2}{2\text{v}}$
$\int\frac{2\text{v}}{1+\text{v}^2}\text{dv}=-\int\frac{\text{dx}}{\text{x}}$
$\log\big|1+\text{v}^2\big|=-\log|\text{x}|+\log|\text{C}|$
$1+\text{v}^2=\frac{\text{C}}{\text{x}}$
$1+\frac{\text{y}^2}{\text{x}^2}=\frac{\text{C}}{\text{x}}$
$\text{x}^2+\text{y}^2=\text{Cx}$
View full question & answer→Question 45 Marks
Represent the following families of curves by forming the corresponding differential equation:
$\text{x}^2+\text{y}^2=\text{ax}^3$
AnswerThe equation of the family of curves is
$\text{x}^2+\text{y}^2=\text{ax}^3\ ...(1)$
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
$2\text{x}+2\text{y}\frac{\text{dy}}{\text{dx}}=3\text{ax}^2$
$\Rightarrow2\text{x}+2\text{y}\frac{\text{dy}}{\text{dx}}=3\Big(\frac{\text{x}^2+\text{y}^2}{\text{x}^3}\Big)\text{x}^2$
$\Rightarrow2\text{x}+2\text{y}\frac{\text{dy}}{\text{dx}}=3\frac{\text{x}^2+\text{y}^2}{\text{x}^3}$
$\Rightarrow2\text{x}^2+2\text{xy}=3\text{x}^2+3\text{y}^2$
$2\text{xy}=\text{x}^2+3\text{y}^2$
It is the required differential equation.
View full question & answer→Question 55 Marks
Form the differential equation of all the circle which pass through the origin and whose centres lies in x-axis.
AnswerThe equation of the family of circles that pass through the origin (0, 0) and whose centres lie on the x-axis is given by
$(\text{x}-\text{a})^2+\text{y}^2=\text{a}^2\ ...(1)$
where a are arbitrary constants.
As this equation has only one arbitrary constant, we shall get a first order differential equation.
Differentiating (1) with respect to x, we get
$2(\text{x}-\text{a})+2\text{y}\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\text{x}-\text{a}+\text{y}\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\text{x}+\text{y}\frac{\text{dy}}{\text{dx}}=\text{a}$
Substituting the value of a in equation (2), we get
$\Big(\text{x}-\text{x}-\text{y}\frac{\text{dy}}{\text{dx}}\Big)^2+\text{y}^2=\Big(\text{x}+\text{y}\frac{\text{dy}}{\text{dx}}\Big)^2$
$\Rightarrow\text{y}^2\Big(\frac{\text{dy}}{\text{dx}}\Big)^2+\text{y}^2=\text{x}^2+2\text{xy}\frac{\text{dy}}{\text{dx}}+\text{y}^2\Big(\frac{\text{dy}}{\text{dx}}\Big)^2$
$\Rightarrow2\text{xy}\frac{\text{dy}}{\text{dx}}+\text{x}^2=\text{y}^2$
It is the required differential equation.
View full question & answer→Question 65 Marks
Solve the following differential equation:$\frac{\text{dy}}{\text{dx}}+\text{y}=\text{e}^{-2\text{x}}$
AnswerHere, $\frac{\text{dy}}{\text{dx}}+\text{y}=\text{e}^{-2\text{x}}$ This is a linear differential equation, comparing it with $\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$ $\text{P}=1,\text{Q}=\text{e}^{-2\text{x}}$ I.F. $=\text{e}^{\int\text{Pdx}}$ $=\text{e}^{\int2\text{dx}}$ $=\text{e}^{\text{x}}$Solution of the equation is given by,
$\text{y}\times(\text{I.F.})=\int\text{Q}\times(\text{I.F.})\text{dx + C}$
$\text{y}\times\text{e}^{\text{x}}=\int\text{e}^{-2\text{x}}\times\text{e}^{\text{x}}\text{dx + C}$ $=\int\text{e}^{-\text{x}}+\text{C}$ $\text{y}\text{e}^{\text{x}}=\frac{\text{e}^{-\text{x}}}{-1}+\text{C}$ $\text{y}=-\text{e}^{-2\text{x}}+\text{C}\text{e}^{-\text{x}}$
View full question & answer→Question 75 Marks
Solve the following differential equations:
$\frac{\text{dy}}{\text{dx}}=2\text{e}^{\text{x}}\text{y}^3,\text{y}(0)=\frac{1}{2}$
AnswerWe have,
$\frac{\text{dy}}{\text{dx}}=2\text{e}^{\text{x}}\text{y}^3,\text{y}(0)=\frac{1}{2}$
$\Rightarrow\frac{1}{\text{y}^3}\text{dy}=2\text{e}^{\text{x}}\text{dx}$
Integrating both sides, we get
$\int\frac{1}{\text{y}^3}\text{dy}=\int2\text{e}^{\text{x}}\text{dx}$
$\Rightarrow-\frac{1}{2\text{y}^3}=2\text{e}^{\text{x}}+\text{C}...(1)$
Given: at $\text{x}=0,\text{y}=\frac{1}{2}$
Substituting the valuse of x and y in (1), we get
$-\frac{1}{2\times\frac{1}{4}}=2\text{e}^{0}+\text{C}$
$\Rightarrow\text{C}=-2-2$
$\Rightarrow\text{C}=-4$
Substituting the value of C in (1), we get
$\Rightarrow-\frac{1}{2\text{y}^2}=2\text{e}^{\text{x}}-4$
$\Rightarrow\text{y}^{2}(8-4\text{e}^{\text{x}})=1$
Hence, $\text{y}^{\text{x}}(8-4\text{e}^{\text{x}})=1$ is the required solution.
View full question & answer→Question 85 Marks
Solve the following differential equation
$\frac{\text{dy}}{\text{dx}}=\text{x}^2+\text{x}-\frac{1}{\text{x}},\text{x}\ne0$
Answer$\frac{\text{dy}}{\text{dx}}=\text{x}^2+\text{x}-\frac{1}{\text{x}}$
$\Rightarrow\text{dy}=\Big(\text{x}^2+\text{x}-\frac{1}{\text{x}}\Big)\text{dx}$
Intergrating both sides, we get
$\Rightarrow\int\text{dy}=\int\Big(\text{x}^2+\text{x}-\frac{1}{\text{x}}\Big)\text{dx}$
$\Rightarrow\text{y}=\frac{\text{x}^3}{3}+\frac{\text{x}^2}{3}-\log|\text{x}|+\text{C}$
Clearly, $\Rightarrow\text{y}=\frac{\text{x}^3}{3}+\frac{\text{x}^2}{3}-\log|\text{x}|+\text{C}$ is defined for all $\text{x}\in\text{R}$ except x = 0
Hence, $\Rightarrow\text{y}=\frac{\text{x}^3}{3}+\frac{\text{x}^2}{3}-\log|\text{x}|+\text{C}$, where $\text{x}\in\text{R}-\{0\},$ is the solution o the given differential equation.
View full question & answer→Question 95 Marks
Show that $\text{y}=\text{A}\cos2\text{x}+\text{B}\sin2\text{x}$ is a solution of the differential equation $\frac{\text{d}^2\text{y}}{\text{dx}^2}+4\text{y}=0$
AnswerWe have,
$\text{y}=\text{A}\cos2\text{x}+\text{B}\sin2\text{x}\ ...(1)$
Differentiating both sides of equation (1) with respect to 3, we get
$\frac{\text{dy}}{\text{dx}}=-2\text{A}\sin2\text{x}-\text{B}\cos2\text{x}\ ...(2)$
Differentiating both sides of equation (2) with respect to 3, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=-4\text{A}\cos2\text{x}+4\text{B}\sin2\text{x}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=-4(\text{A}\cos2\text{x}+4\text{B}\sin2\text{x})$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=-4\text{y}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}+4\text{y}=0$
Hence, the given function is the solution to the given differential equation.
View full question & answer→Question 105 Marks
Show that $\text{y}=\text{ae}^{2\text{x}}+\text{be}^{-\text{x}}$ is a solution of the differential equation $\frac{\text{d}^2\text{y}}{\text{dx}^2}-\frac{\text{dy}}{\text{dx}}-2\text{y}=0$
AnswerWe have,
$\text{y}=\text{ae}^{2\text{x}}+\text{be}^{-\text{x}}\ ...(1)$
Differentiating both sides of equation (1) with respect to 3, we get
$\frac{\text{dy}}{\text{dx}}=\text{ae}^{2\text{x}}+\text{be}^{-\text{x}}\ ...(2)$
Differentiating both sides of equation (2) with respect to 3, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=4\text{ae}^{2\text{x}}+\text{be}^{-\text{x}}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=2\text{ae}^{2\text{x}}-\text{be}^{-\text{x}}+2\text{ae}^{2\text{x}}+2\text{be}^{-\text{x}}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\big(2\text{ae}^{2\text{x}}-\text{be}^{-\text{x}}\big)+2\big(\text{ae}^{2\text{x}}+2\text{be}^{-\text{x}}\big)$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{dy}}{\text{dx}}+2\text{y}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}-\frac{\text{dy}}{\text{dx}}-2\text{y}=0$
Hence, the given function is the solution to the given differential equation.
View full question & answer→Question 115 Marks
Solve the following differential equations:
$\frac{\text{dy}}{\text{dx}}=\frac{\text{x + y}}{\text{x}-\text{y}}$
AnswerWe have,$\frac{\text{dy}}{\text{dx}}=\frac{\text{x + y}}{\text{x}-\text{y}}$
This is a homogeneous differential equation.
Put y = vx and $\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}}$, we get
$\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{{\text{x}}+\text{vx}}{\text{x}-\text{vx}}$
$\Rightarrow\ \text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{1+\text{v}}{1-\text{v}}$
$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{1+\text{v}}{1-\text{v}}-\text{v}$
$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{1+\text{v}^2}{1-\text{v}}$
$\frac{1-\text{v}}{1+\text{v}^2}\text{dv}=\frac{1}{\text{x}}\text{dx}$
Integrating both sides, we get
$\int\frac{1-\text{v}}{1+\text{v}^2}\text{dv}=\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\ \int\frac{1}{1+\text{v}^2}\text{dv}-\int\frac{\text{v}}{1+\text{v}^2}\text{dv}=\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\ \int\frac{1}{1+\text{v}^2}\text{dv}-\frac{1}2\int\frac{2\text{v}}{1+\text{v}^2}\text{dv}=\int\frac{1}{\text{x}}\text{dx}$
$\tan^{-1}\text{v}-\frac{1}2\log\big|1+\text{v}^2\big|=\log|\text{x}|+\text{C}$
Putting $\text{v}=\frac{\text{y}}{\text{x}}$, we get
$\Rightarrow\ \tan^{-1}\Big(\frac{\text{y}}{\text{x}}\Big)-\frac{1}2\log\Big|1+\frac{\text{y}^2}{\text{x}^2}\Big|=\log|\text{x}|+\text{C}$
$\Rightarrow\ \tan^{-1}\Big(\frac{\text{y}}{\text{x}}\Big)=\frac{1}2\log\Big|1+\frac{\text{y}^2}{\text{x}^2}\Big|+\log|\text{x}|+\text{C}$
$\Rightarrow\ \tan^{-1}\Big(\frac{\text{y}}{\text{x}}\Big)=\frac{1}2\log\Big|\frac{\text{x}^2+\text{y}^2}{\text{x}^2}\Big|+\log|\text{x}|+\text{C}$
$\Rightarrow\ \tan^{-1}\Big(\frac{\text{y}}{\text{x}}\Big)=\frac{1}2\log\Big|\text{x}^2+\text{y}^2\Big|-\frac{1}2\log|\text{x}^2|+\log|\text{x}|+\text{C}$
$\Rightarrow\ \tan^{-1}\Big(\frac{\text{y}}{\text{x}}\Big)=\frac{1}2\log\Big|\text{x}^2+\text{y}^2\Big|-\log|\text{x}|+\log|\text{x}|+\text{C}$
$\Rightarrow\ \tan^{-1}\Big(\frac{\text{y}}{\text{x}}\Big)=\frac{1}2\log\Big|\text{x}^2+\text{y}^2\Big|+\text{C}$
Hence, $\tan^{-1}\Big(\frac{\text{y}}{\text{x}}\Big)=\frac{1}2\log\Big|\text{x}^2+\text{y}^2\Big|+\text{C}$ is the required solution.
View full question & answer→Question 125 Marks
Show that $\text{y}=\frac{\text{a}}{\text{x}}+\text{b}$ is a solution of the differential equation $\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{2}{\text{x}}\Big(\frac{\text{dy}}{\text{dx}}\Big)=0$
AnswerWe have,
$\text{y}=\frac{\text{a}}{\text{x}}+\text{b}\ ...(1)$
Differentiating both sides of equation (1) with respect to 3, we get
$\frac{\text{dy}}{\text{dx}}=-\frac{\text{a}}{\text{x}^2}\ ...(2)$
Differentiating both sides of equation (2) with respect to 3, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=2\frac{\text{a}}{\text{x}^3}$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\frac{2}{\text{x}}\Big(-\frac{\text{a}}{\text{x}^2}\Big)$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\frac{2}{\text{x}}\Big(\frac{\text{dy}}{\text{dx}}\Big)$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}+\frac{2}{\text{x}}\Big(\frac{\text{dy}}{\text{dx}}\Big)=0$
Hence, the given function is the solution to the given differential equation.
View full question & answer→Question 135 Marks
Solve the following differential equation
$\text{C}(\text{x})=2+0.15\text{x},\text{C}(0)=100$
Answer $\text{C}(\text{x})=2+0.15\text{x},\text{C}(0)=100$$\text{C}'(\text{x})\text{dx}=(2+0.15\text{x})\text{dx}$
$\int\text{C}'(\text{x})\text{dx}=\int2\text{dx}+0.15\int\text{x dx}$
$\text{C}(\text{x})=2\text{x}+0.15\frac{\text{x}^2}{2}+\text{C}\ ...(1)$
Put x = 0, c(x) = 100
100 = 2(0) + 0 + c
100 = c
Put c = 100 in equation 1
$\text{c}(\text{x})=2\text{x}+(0.15)\frac{\text{x}^2}{2}+100$
View full question & answer→Question 145 Marks
Solve the following differential equation
$\frac{\text{dy}}{\text{dx}}=\text{x}^5\tan^{-1}(\text{x}^3)$
Answer$\frac{\text{dy}}{\text{dx}}=\text{x}^5\tan^{-1}(\text{x}^3)$$\text{dy}=\text{x}^5\tan^{-1}(\text{x}^3)\text{dx}$
$\int\text{dy}=\int\text{x}^5\tan^{-1}(\text{x}^3)\text{dx}$
put $\text{x}^3=\text{t}$
$\Rightarrow3\text{x}^2\text{dx}=\text{dt}$
$\Rightarrow\text{x}^2\text{dx}=\frac{\text{dt}}{3}$
So,
$\int\text{dy}=\frac{1}{3}\Big[\tan^{-1}\text{t}\int\text{t dt}=\int\Big(\frac{1}{1+\text{t}^2}\Big)\times\int\text{t dx}\Big)\Big]\text{dt}+\text{C}$
Using integration by parts
$\text{y}=\frac{1}{3}\Big[\frac{\text{t}^2}{2}+\tan^{-1}-\int\frac{\text{t}^2}{2(\text{t}^2+1)}\text{dt}\Big]+\text{C}$
$=\frac{1}{6}\text{t}^2\tan^{-1}\text{t}-\frac{1}{6}\int\Big(\frac{\text{t}^2}{\text{t}^2+1}\Big)\text{dt}+\text{C}$
$\text{y}=\frac{1}{6}\text{t}^2\tan^{-1}\text{t}-\frac{1}{6}\int\Big(1-\frac{1}{\text{t}^2+1}\Big)\text{dt}+\text{C}$
$=\frac{1}{6}\text{t}^2\tan^{-1}\text{t}-\frac{1}{6}\text{t}+\frac{1}{6}\tan^{-1}\text{t}+\text{C}$
$\text{y}=\frac{1}{6}(\text{t}^2+1)\tan^{-1}\text{t}-\frac{1}{6}\text{t}+\text{C}$
$\text{y}=\frac{1}{6}[(\text{t}^2+1)\tan^{-1}\text{t}-\text{t}]+\text{C}$
So,
$\text{y}=\frac{1}{6}[(\text{x}^6+1)\tan^{-1}(\text{x}^3)-\text{x}^3]+\text{C}$
View full question & answer→Question 155 Marks
Assume that a rain drop evaporates at a rate proportional to its surface area. Form a differential equation involving the rate of change of the radius of the rain drop.
AnswerLet A be the surface area of rain drain, v be its volume, and r be the radius of rain drop.
Given:
$\frac{\text{dV}}{\text{dt}}=\text{A}$
$\frac{\text{dV}}{\text{dt}}=-\text{KA}$ [negative because V decrease with increase in t]
Where K is the constant of proportionality.
So,
$\frac{\text{d}}{\text{dt}}\Big(\frac{4\pi}{3}\text{r}^3\Big)=-\text{K}(4\pi\text{r}^2)$
$4\pi\text{r}^2\frac{\text{dr}}{\text{dt}}=-\text{K}(4\pi\text{r}^2)$
$\frac{\text{dr}}{\text{dt}}=-\text{K}$
View full question & answer→Question 165 Marks
Solve the following differential equation
$(\sin\text{x}+\cos\text{x})\text{dy}+(\cos\text{x}+\sin\text{x})\text{dx}=0$
Answer We have,$(\sin\text{x}+\cos\text{x})\text{dy}+(\cos\text{x}+\sin\text{x})\text{dx}=0$
$\Rightarrow{\text{dy}}=-\Big(\frac{\cos\text{x}-\sin\text{x}}{\sin\text{x}-\cos\text{x}}\Big)\text{dx}$
Integrating both sides, we get
$\int{\text{dy}}=-\int\Big(\frac{\cos\text{x}-\sin\text{x}}{\sin\text{x}-\cos\text{x}}\Big)\text{dx}$
$\Rightarrow\text{y}=-\int\Big(\frac{\cos\text{x}-\sin\text{x}}{\sin\text{x}-\cos\text{x}}\Big)\text{dx}$
Putting $\sin\text{x}+\cos\text{x}=\text{t}$
$\Rightarrow(\cos\text{x}-\sin\text{x})\text{dx}=\text{dt}$
$\therefore\text{y}=-\int\frac{\text{dt}}{\text{t}}$
$\Rightarrow\text{y}=-\log|\text{t}|+\text{C}$
$\Rightarrow\text{y}=-\log|\sin\text{x}+\cos\text{x}|+\text{C}$
$\Rightarrow\text{y}=+\log|\sin\text{x}+\cos\text{x}|=\text{C}$
Hence, $\text{y}=+\log|\sin\text{x}+\cos\text{x}|=\text{C}$ is the solution to the given differential equation.
View full question & answer→Question 175 Marks
Solve the following differential equation
$\frac{\text{dy}}{\text{dx}}=\frac{\text{x e }^\text{x}\log\text{x}+\text{e}^\text{x}}{\text{x}\cos\text{y}}$
AnswerWe have,$\frac{\text{dy}}{\text{dx}}=\frac{\text{x e }^\text{x}\log\text{x}+\text{e}^\text{x}}{\text{x}\cos\text{y}}$
$\Rightarrow\text{x}\cos\text{y dy}=(\text{x e}^\text{x}\log\text{x}+\text{e}^\text{x})\ \text{dx}$
$\Rightarrow\cos\text{y dy}=\Big(\text{e}^\text{x}\log\text{x}+\frac{1}{\text{x}}\text{e}^\text{x}\Big)\ \text{dx}$
Integrating both sides, we get
$\int\cos\text{y dy}=\int\Big(\text{e}^\text{x}\log\text{x}+\frac{1}{\text{x}}\text{e}^\text{x}\Big)\text{dx}$
$\Rightarrow\sin\text{y}=\log\text{x}\int\text{e} ^\text{x}\text{dx}-\int\frac{1}{\text{x}}\text{e}^\text{x}\text{dx}+\int\frac{1}{\text{x}}\text{e}^\text{x}\text{dx}$
$\Rightarrow\sin\text{y}=\text{e}^\text{x}\log\text{x}+\text{C}$
Hence, $\sin\text{y}=\text{e}^\text{x}\log\text{x}+\text{C}$ is the required solution.
View full question & answer→Question 185 Marks
Show that $\text{y}=\text{A}\cos\text{x}+\text{B}\sin\text{x}$ is a solution of the differential equation $\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{y}=0$
AnswerWe have,
$\text{y}=\text{A}\cos\text{x}+\text{B}\sin\text{x}\ ...(1)$
Differentiating both sides of equation (1) with respect to 3, we get
$\frac{\text{dy}}{\text{dx}}=\text{A}\cos\text{x}+\text{B}\sin\text{x}\ ...(2)$
Differentiating both sides of equation (2) with respect to 3, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\text{A}\cos\text{x}-\text{B}\sin\text{x}$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=(\text{A}\cos\text{x}+\text{B}\sin\text{x})$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\text{y}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{y}=0$
Hence, the given function is the solution to the given differential equation.
View full question & answer→Question 195 Marks
show that the differential equation of which $\text{y}=2(\text{x}^2-1)+\text{ce}^{-\text{x}^{2}}$ is a solution, is $\frac{\text{dy}}{\text{dx}}+2\text{xy}=4\text{x}^3$
AnswerThe given equation is
$\text{y}=2(\text{x}^2-1)+\text{ce}^{-\text{x}^{2}} ...(1)$
Where c is a parameter.
As this equation has one arbitrary constant, we shall get a differential equation of first order.
Differentiating equation (1) with respect to x, we get
$\frac{\text{dy}}{\text{dx}}=2(2\text{x})+\text{ce}^{-\text{x}^{2}}(-2\text{x})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=4\text{x}-2\text{xce}^{-\text{x}^{2}}$
From (1) and (2), we get
$\frac{\text{dy}}{\text{dx}}=4\text{x}-2\text{x}[\text{y}-2\text{x}^2+2]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=4\text{x}-2\text{xy}+4\text{x}^3-4\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}+2\text{xy}=4\text{x}^3$
Hence, $\text{y}=2(\text{x}^2-1)+\text{ce}^{-\text{x}^{2}}$ is the solution to the differential equation $\frac{\text{dy}}{\text{dx}}+2\text{xy}=4\text{x}^3.$
View full question & answer→Question 205 Marks
Solve the following differential equation $(1+\text{x}^2)\text{dy}=\text{xy dx}$
AnswerWe have $(1+\text{x}^2)\text{dy}=\text{xy dx}$
$\Rightarrow\frac{1}{\text{y}}\text{dy}=\frac{\text{x}}{1+\text{x}^2}\ \text{dx}$
Integrating both sides, we get
$\int\frac{1}{\text{y}}\text{dy}=\int\frac{\text{x}}{1+\text{x}^2}\ \text{dx}$
Substituting $1+ x^2 = t,$ we get
$2\text{x dx}=\text{dt}$
$\therefore\int\frac{1}{\text{y}}\text{dy}=\frac{1}{2}\int\frac{1}{\text{t}}\text{dt}$
$\Rightarrow\log|\text{y}|=\frac{1}{2}\log|\text{t}|+\log\text{C}$
$\Rightarrow\log|\text{y}|=\frac{1}{2}\log|1+\text{x}^2|+\log\text{C}$
$\Rightarrow\log|\text{y}|=\log\Big[\text{C}\sqrt{1+\text{x}^2}\Big]$
$\Rightarrow\text{y}=\text{C}\sqrt{1+\text{x}^2}$
Hence $,\text{y}=\text{C}\sqrt{1+\text{x}^2}$ is the required solution.
View full question & answer→Question 215 Marks
The slope of a curve at each of its points is equal to the square of the abscissa of the point. Find the particular curve through the point $(−1, 1)$.
AnswerGiven,
Slope of tangent at $(x, y) = x^2$
$\frac{\text{dy}}{\text{dx}}=\text{x}^{2}$
$\text{dy}=\text{x}^{2}\text{dx}$
$\int \text{dy}=\int\text{x}^{2}\text{dx}$
$\text{y}=\frac{\text{x}^{3}}{3}+\text{C}\ ...(\text{i})$
It is passing through $(-1, 1)$
$1=\frac{(-1)}{3}+\text{C}$
$1=-\frac{1}{3}+\text{C}$
$\text{C}=\frac{4}{3}$
Put is equation,
$\text{y}=\frac{\text{x}^{3}}{3}+\frac{4}{3}$
$3\text{y}=\text{x}^{3}+{4}$
View full question & answer→Question 225 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 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
$8\text{x}-4\text{a}-2\text{y}\frac{\text{dy}}{\text{dx}}=0\ ...(2)$
$\Rightarrow-\text{y}\frac{\text{dy}}{\text{dx}}+4\text{x}=2\text{a}$
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 245 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 255 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 265 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 275 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 $\text{m}+3=0$
View full question & answer→Question 285 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 295 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 305 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 315 Marks
Form the differential equation of the family of curves represented by the equation (a being the perimeter):$(\text{x}-\text{a})^2+2\text{y}^2=\text{a}^2$
AnswerThe equation of the family of curves is
$(\text{x}-\text{a})^2+2\text{y}^2=\text{a}^2\ ...(1)$
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
$2\text{x}-2\text{a}+4\text{y}\frac{\text{dy}}{\text{dx}}=0\ ...(2)$
Now, from (1), we get
$2\text{a}=\frac{\text{x}^2-2\text{y}^2}{\text{x}}\ ...(3)$
From (2) and (3), we get
$2\text{x}-\frac{\text{x}^2+2\text{y}^2}{\text{x}}+4\text{y}\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow2\text{x}^2-\text{x}^2-2\text{y}^2+4\text{xy}\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow4\text{xy}\frac{\text{dy}}{\text{dx}}+\text{x}^2-2\text{y}^2=0$
$\Rightarrow4\text{xy}\frac{\text{dy}}{\text{dx}}=2\text{y}^2-\text{x}^2$
It is the required differential equation.
View full question & answer→Question 325 Marks
Solve the following differential equations:$\frac{\text{dy}}{\text{dx}}=\text{y}\tan\text{ x, y}(0)=1$
Answer$\frac{\text{dy}}{\text{dx}}=\text{y}\tan\text{ x, y}(0)=1$
$\Rightarrow\frac{1}{\text{y}}\text{dy}=\tan\text{ x dx}$
Integrating both sides, we get
$\int \frac{1}{\text{y}}\text{dy}=\int\tan\text{ x dx}$
$\Rightarrow\log|\text{y}|=\log|\sec\text{x}|+\text{C}...(1)$
We know that at $\text{x}=0$ and $\text{y}=1.$
Substituting the values of x and y in (1), we get
$\log|1|=\log|1|+\text{C}$
$\Rightarrow\text{C}=0$
substituting the value of C in (1), we get
$\log|\text{y}|=\log|\sec\text{x}|+0$
$\Rightarrow\text{y}=\sec\text{x}$
Hence, $\text{y}=\sec\text{x},$ where $\text{x}\in\Big(\frac{-\pi}{2},\frac{\pi}{2}\Big),$ is the required solution.
View full question & answer→Question 335 Marks
Solve the following differential equations:$\frac{\text{dy}}{\text{dx}}=\text{y}\tan2\text{x, y}(0)=2$
AnswerWe have,
$\frac{\text{dy}}{\text{dx}}=\text{y}\tan2\text{x, y}(0)=2$
$\Rightarrow\frac{1}{\text{y}}\text{dy}=\tan2\text{x dx}$
Integrating both sides, we get
$\int\frac{1}{\text{y}}\text{dy}=\int\tan2\text{x dx}$
$\Rightarrow\log|\text{y}|=\frac{1}{2}\log|\sec2\text{x}|+\frac{1}{2}\log\text{C}$
$\Rightarrow\text{y}^2=\text{C}\sec2\text{x}\dots(1)$
It is given that at $\text{x}=0,\text{y}=2.$
$\therefore\text{C}=4$
Substituting the value of C in (1), we get
$\therefore\text{y}^2=\frac{4}{\cos2\text{x}}$
$\Rightarrow\text{y}=\frac{2}{\sqrt{\cos2\text{x}}}$
Hence, $\text{y}=\frac{2}{\sqrt{\cos2\text{x}}}$ is the required solution.
View full question & answer→Question 345 Marks
Find one-parameter families of solution curves of the following differential equation: (or solve the following differential equation)$(\text{x}\log\text{x})\frac{\text{dy}}{\text{dx}}+\text{y}=\log\text{x}$
AnswerHere, $(\text{x}\log\text{x})\frac{\text{dy}}{\text{dx}}+\text{y}=\log\text{x}$ $\frac{\text{dy}}{\text{dx}}+\frac{\text{y}}{\text{x}\log\text{x}}=\frac{1}{\text{x}}$ It is a linear differential equation.Comparing it with, $\frac{\text{dy}}{\text{dx}}+\text{Py = Q}$ $\text{P}=\frac{1}{\text{x}\log\text{x}},\text{Q}=\frac{1}{\text{x}}$ I.F. $=\text{e}^{\int\text{Pdx}}$ $=\text{e}^{\int\frac{1}{\text{x}\log\text{x}}\text{dx}}$ $=\text{e}^{\log|\log\text{x}|}$ $=\log\text{x}$Solution of the equaion is given by,
$\text{y}\times(\text{I.F.})=\int\text{Q}\times(\text{I.F.})\text{dx + C}$ $\text{y}(\log\text{x})=\int\frac{1}{\text{x}}(\log\text{x})\text{dx + C}$ $\text{y}(\log\text{x})=\frac{(\log\text{x})^2}{2}+\text{C}$ $\text{y}=\frac{1}{2}\log\text{x}+\frac{\text{C}}{\log\text{x}},\text{x}>0,\text{x}\neq1$
View full question & answer→Question 355 Marks
Solve the following differential equations:$\text{cosec x}\log\text{y}\frac{\text{dy}}{\text{dx}}+\text{x}^2\text{y}^2=0$
Answer$\text{cosec x}\log\text{y}\frac{\text{dy}}{\text{dx}}+\text{x}^2\text{y}^2=0$
$\Rightarrow\text{cosec x }\log\text{y}\frac{\text{dy}}{\text{dx}}=-\text{x}^2\text{y}^2$
$\Rightarrow\frac{1}{\text{y}^2}\log\text{ y dy}=-\frac{\text{x}^2}{\text{cosec x}}\text{dx}$
$\Rightarrow\frac{1}{\text{y}^2}\log\text{ y dy}=-\text{x}^2\sin\text{x dx}$
$\Rightarrow\int\frac{1}{\text{y}^2}\log\text{ y dy}=-\int\text{x}^2\sin\text{x dx}$
$\Rightarrow-\frac{\log\text{y}}{\text{y}}+\int\frac{1}{\text{y}}\times\frac{1}{\text{y}}=-\Big[-\text{x}^2\cos\text{x}+\int2\text{x}\cos\text{x dx}\Big]+\text{C}$
$\Rightarrow-\frac{\log\text{y}}{\text{y}}-\frac{1}{\text{y}}=-\Big[-\text{x}^2\cos\text{x}+2\text{x}\sin\text{x}-2\int\sin\text{x dx}\Big]+\text{C}$
$\Rightarrow-\Big(\frac{1+\log\text{y}}{\text{y}}\Big)=-\big[-\text{x}^2\cos\text{x}+2\text{x}\sin\text{x}+2\cos\text{x dx}\big]+\text{C}$
$\Rightarrow-\Big(\frac{1+\log\text{y}}{\text{y}}\Big)-\text{x}^2\cos\text{x}+2(\text{x}\sin\text{x}+\cos\text{x})=\text{C}$
View full question & answer→Question 365 Marks
Solve the following differential equations:$(1+\text{x})(1+\text{y}^2)\text{dx}+(1+\text{y})(1+\text{x}^2)\text{dy}=0$
AnswerWe have,
$(1+\text{x})(1+\text{y}^2)\text{dx}+(1+\text{y})(1+\text{x}^2)\text{dy}=0$
$\Rightarrow(1+\text{x})(1+\text{y}^2)\text{dx}=-(1+\text{y})(1+\text{x}^2)\text{dy}$
$\Rightarrow\frac{1+\text{x}}{1+\text{x}^2}\text{dx}=-\frac{1+\text{y}}{1+\text{y}^2}\text{dy}$
Integrating both sides, we get
$\Rightarrow\frac{1+\text{x}}{1+\text{x}^2}\text{dx}=-\frac{1+\text{y}}{1+\text{y}^2}\text{dy}$
$\Rightarrow\int\frac{1}{1+\text{x}^2}\text{dx}+\int\frac{\text{x}}{1+\text{x}^2}\text{dx}=-\int\frac{1}{1+\text{y}^2}\text{dy}-\int\frac{\text{y}}{1+\text{y}^2}\text{dy}$
Substituting $1+\text{x}^2=\text{t}$ in the second integral of LHS and $1+\text{y}^2=\text{u}$ in the second integral of RHS, we get
$2\text{x dx = dt}$ and $2\text{y dy = du}$
$\therefore\int \frac{1}{1+\text{x}^2}\text{dx}+\frac{1}{2}\int\text{dt}=-\int\frac{1}{1+\text{y}^2}\text{dy}-\frac{1}{2}\int\frac{1}{\text{u}}\text{du}$
$\Rightarrow\tan^{-1}\text{x}+\frac{1}{2}\log|\text{t}|=-\tan^{-1}\text{y}-\frac{1}{2}\log|\text{u}|+\text{C}$
$\Rightarrow\tan^{-1}\text{x}+\frac{1}{2}\log|1+\text{x}^2|=-\tan^{-1}\text{y}-\frac{1}{2}\log|1+\text{y}^2|+\text{C}$
$\Rightarrow\tan^{-1}\text{x}+\tan^{-1}\text{y}+\frac{1}{2}\log|1+\text{x}^2|+\frac{1}{2}\log|1+\text{y}^2|=\text{C}$
$\Rightarrow\tan^{-1}\text{x}+\tan^{-1}\text{y}+\frac{1}{2}\log\big|(1+\text{x}^2)(1+\text{y}^2)\big|=\text{C}$
Hence, $\tan^{-1}\text{x}+\tan^{-1}\text{y}+\frac{1}{2}\log\big|(1+\text{x}^2)(1+\text{y}^2)\big|=\text{C}$ is the required solution.
View full question & answer→Question 375 Marks
Solve the following differential equation:$\frac{\text{dy}}{\text{dx}}+\frac{4\text{x}}{\text{x}^2+1}\text{y}+\frac{1}{(\text{x}^2+1)^2}=0$
AnswerWe have,
$\frac{\text{dy}}{\text{dx}}+\frac{4\text{x}}{\text{x}^2+1}\text{y}+\frac{1}{(\text{x}^2+1)^2}=0$
$\frac{\text{dy}}{\text{dx}}+\frac{4\text{x}}{\text{x}^2+1}\text{y}=-\frac{1}{(\text{x}^2+1)^2}\ \dots(1)$
Clearly, it is a linear differential equation of the form
$\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
where
$\text{P}=\frac{4\text{x}}{\text{x}^2+1}$
$\text{Q}=-\frac{1}{(\text{x}^2+1)^2}$
$\therefore I.F. =\text{e}^{\int\text{Pdx}}$
$=\text{e}^{2\int\frac{2\text{x}}{\text{x}^2+1}\text{dx}}$
$=\text{e}^{2\log|\text{x}^2+1|}$
$=(\text{x}^2+1)^2$
Multiplying both sides of $(1)$ by $(x^2 + 1)^2,$ we get
$(\text{x}^2+1)^2\Big(\frac{\text{dy}}{\text{dx}}+\frac{4\text{x}}{\text{x}^2+1}\text{y}\Big)=(\text{x}^2+1)^2\Big[-\frac{1}{(\text{x}^2+1)^2}\Big]$
$\Rightarrow\ (\text{x}^2+1)^2\frac{\text{dy}}{\text{dx}}+4\text{x}(\text{x}^2+1)\text{y}=-1$
Integrating both sides with respect to $x,$ we get
$(\text{x}^2+1)^2\text{y}=-\int\text{dx + C}$
$\Rightarrow\ (\text{x}^2+1)^2\text{y}=-\text{x + C}$
Hence, $(\text{x}^2+1)^2\text{y}=-\text{x + C}$ is the required solution.
View full question & answer→Question 385 Marks
Solve the following differential equations:$\text{y}(1-\text{x}^2)\frac{\text{dy}}{\text{dx}}=\text{x}(1+\text{y}^2)$
AnswerWe have,
$\text{y}(1-\text{x}^2)\frac{\text{dy}}{\text{dx}}=\text{x}(1+\text{y}^2)$
$\Rightarrow\frac{\text{y}}{1+\text{y}^2}\text{dy}=\frac{\text{x}}{1-\text{x}^2}\text{dx}$
Integrating both sides,
$\int\frac{\text{y}}{1+\text{y}^2}\text{dy}=\int\frac{\text{x}}{1-\text{x}^2}\text{dx}$
Substituting $1+\text{y}^2=\text{t}$ and $1-\text{x}^2=\text{u}$
$2\text{ydy = dt}$ and $-2\text{x dx = du}$
$\therefore\frac{1}{2}\int\frac{1}{\text{t}}=\frac{-1}{2}\int\frac{1}{\text{u}}\text{du}$
$\Rightarrow\frac{1}2{}\log|\text{t}|=-\frac{1}{2}\log|\text{u}|+\log\text{C}$
$\Rightarrow\frac{1}{2}|1+\text{y}^2|=-\frac{1}{2}\log|1-\text{x}^2|+\log\text{C}$
$\Rightarrow\frac{1}{2}\big[\log|1+\text{y}^2|+\log|1-\text{x}^2|\big]=\log\text{C}$
$\Rightarrow\log(|1+\text{y}^2||1-\text{x}^2|)=2\log\text{C}$
$\Rightarrow(1+\text{y}^2)(1-\text{x}^2)=\text{C}^2$
$\Rightarrow(1+\text{y}^2)(1-\text{x}^2)=\text{C}_1,$ where $\text{C}_1=\text{C}^2$
Hence, $(1+\text{y}^2)(1-\text{x}^2)=\text{C}_1$ is the required solution.
View full question & answer→Question 395 Marks
Solve the following differential equation:$\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}=\text{x}\text{e}^{\text{x}}$
AnswerHere, $\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}=\text{x}\text{e}^{\text{x}}$ $\frac{\text{dy}}{\text{dx}}+\frac{\text{y}}{\text{x}}=\text{e}^{\text{x}}$ It is a linear differential equation, comparing it with $\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$ $\text{P}=\frac{1}{\text{x}}, \text{Q}=\text{e}^{\text{x}}$ I.F. $=\text{e}^{\int\text{Pdx}}$ $=\text{e}^{\int\frac{1}{\text{x}}\text{dx}}$ $=\text{e}^{\log\text{x}}$ $=\text{x}$Solution of the equation is given by,
$\text{y}\times(\text{I.F.})=\int\text{Q}\times(\text{I.F.})\text{dx + C}$ $\text{y}\times(\text{x})=\int\text{e}^{\text{x}}\times\text{xdx + C}$ $\text{xy}=\text{x}\int\text{e}^{\text{x}}\text{dx}-\int(1\times\int\text{e}^{\text{x}}\text{dx})\text{dx + C}$ Using integration by parts $=\text{x}\text{e}^{\text{x}}-\int\text{e}^{\text{x}}\text{dx}+\text{C}$ $=\text{x}\text{e}^{\text{x}}-\text{e}^{\text{x}}+\text{C}$ $\text{xy}=(\text{x}-1)\text{e}^{\text{x}}+\text{C}$ $\text{y}=\Big(\frac{\text{x}-1}{\text{x}}\Big)\text{e}^{\text{x}}+\frac{\text{C}}{\text{x}},\text{x}>0$
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)$\frac{\text{dy}}{\text{dx}}-\text{y}=\cos2\text{x}$
AnswerHere,
$\frac{\text{dy}}{\text{dx}}-\text{y}=\cos2\text{x}$
It is a linear differential equation. Comparing it with,
$\frac{\text{dy}}{\text{dx}}+\text{Py = Q}$
$\text{P}=-1,\text{Q}=\cos2\text{x}$
I.F. $=\text{e}^{\int\text{Pdx}}$
$=\text{e}^{-\int\text{dx}}$
$=\text{e}^{-\text{x}}$
Solution of the equation is given by,
$\text{y}\times(\text{I.F.})=\int\text{Q}\times(\text{I.F.})\text{dx + C}$
$\text{y}\times\text{e}^{-\text{x}}=\int\cos2\text{x}\times\text{e}^{-\text{x}}\text{dx + C}\ \dots(\text{i})$
$\text{I}=\int\cos2\text{x}\text{e}^{-\text{x}}\text{dx}=\cos2\text{x}\times(-\text{e}^{-\text{x}})-\int\Big(\frac{\sin2\text{x}}{2}\Big)\text{e}^{-\text{x}}\text{dx}$ [Using integration by parts]
$\text{I}=-\text{e}^{-\text{x}}\cos2\text{x}-\frac{1}2\Big[\big(-\sin2\text{x}\text{e}^{-\text{x}}\big)+\int\frac{\cos2\text{x}}{2}\text{e}^{-\text{x}}\text{dx}\Big]$
$\text{I}=-\text{e}^{-\text{x}}\cos2\text{x}+\frac{1}2\sin2\text{x}\text{e}^{-\text{x}}-\frac{1}4\text{I}$
$\frac{5}4\text{I}=\frac{\text{e}^{-\text{x}}}{2}(\sin2\text{x}-2\cos2\text{x})$
$\text{I}=\frac{2}5\text{e}^{-\text{x}}(\sin2\text{x}-2\cos2\text{x})$
So, solution of the equation is given by
$\text{y}=\frac{2}5(\sin2\text{x}-2\cos2\text{x})+\text{C}\text{e}^{-\text{x}}$
View full question & answer→Question 415 Marks
Solve the following differential equations:$\cos\text{y}\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}},\text{y}(0)=\frac{\pi}{2}$
Answer$\cos\text{y}\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}},\text{y}(0)=\frac{\pi}{2}$
$\Rightarrow\cos\text{y dy = e}^{\text{x}}\text{ dx}$
Integrating both sides, we get
$\int\cos\text{y dy}=\int\text{e}^{\text{x}}\text{ dx}$
$\Rightarrow\sin\text{y}=\text{e}^{\text{x}}+\text{C}...(1)$
We know that at $\text{x}=0,\text{y}=\frac{\pi}{2}.$
Substituting the valuse of x and y in (1), we get
$1=1+\text{C}$
$\Rightarrow\text{C}=0$
Substituting the value of C in (1), we get
$\sin\text{y}=\text{e}^{\text{x}}$
$\Rightarrow\text{y}=\sin^{-1}(\text{e}^{\text{x}})$
Hence, $\text{y}=\sin^{-1}(\text{e}^{\text{x}})$ is the required solution.
View full question & answer→Question 425 Marks
Solve the following differential equations:$\frac{\text{dy}}{\text{dx}}=2\text{e}^{2\text{x}}\text{y}^2,\text{y}(0)=-1$
Answer$\frac{\text{dy}}{\text{dx}}=2\text{e}^{2\text{x}}\text{y}^2,\text{y}(0)=-1$
$\Rightarrow\frac{1}{\text{y}^2}\text{dy}=2\text{e}^{2\text{x}}\text{dx}$
Integrating both sides, we get
$\int\frac{1}{\text{y}^2}\text{dy}=2\int\text{e}^{2\text{x}}\text{dx}$
$\Rightarrow\frac{-1}{\text{y}}=\text{e}^{2\text{x}}+\text{C}...(1)$
We know that at $\text{x}=0,\text{y}=-1.$
Substituting the values of x and y in (1), we get
$1=1+\text{C}$
$\Rightarrow\text{C}=0$
Substituting the value of C in (1), we get
$-\frac{1}{\text{y}}=\text{e}^{2\text{x}}$
$\Rightarrow\text{y}=-\text{e}^{-2\text{x}}$
Hence, $\text{y}=-\text{e}^{-2\text{x}}$ is the required soluton.
View full question & answer→Question 435 Marks
Solve the following differential equation:$\frac{\text{dy}}{\text{dx}}+2\text{y}=4\text{x}$
AnswerWe have,
$\frac{\text{dy}}{\text{dx}}+2\text{y}=4\text{x}\ \dots(1)$
Clearly, it is a linear differential equation of the form
$\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
where
$\text{P}=2$
$\text{Q}=4\text{x}$
$\therefore$ I.F. $=\text{e}^{\int\text{Pdx}}$
$=\text{e}^{\int2\text{dx}}$
$=\text{e}^{2\text{x}}$
Multiplying both sides of (1) by $\text{e}^{2\text{x}},$ we get
$\text{e}^{2\text{x}}\Big(\frac{\text{dy}}{\text{dx}}+2\text{y}\Big)=\text{e}^{2\text{x}}4\text{x}$
$\Rightarrow\ \text{e}^{2\text{x}}\frac{\text{dy}}{\text{dx}}+2\text{e}^{2\text{x}}\text{y}=\text{e}^{2\text{x}}4\text{x}$
Integrating both sides with respect to x, we get
$\text{y}\text{e}^{2\text{x}}=4\int\text{x}\text{e}^{2\text{x}}\text{dx + C}$
$\Rightarrow\ \text{y}\text{e}^{2\text{x}}=4\int\text{x}\text{e}^{2\text{x}}\text{dx + C}$
$\Rightarrow\ \text{y}\text{e}^{2\text{x}}=4\text{x}\int\text{e}^{2\text{x}}\text{dx}-4\int\Big[\frac{\text{d}}{\text{dx}}(\text{x})\int\text{e}^{2\text{x}}\text{dx}\Big]\text{dx + C}$
$\Rightarrow\ \text{y}\text{e}^{2\text{x}}=4\text{x}\frac{\text{e}^{2\text{x}}}2-4\times\frac{1}2\int\text{e}^{2\text{x}}\text{dx + C}$
$\Rightarrow\ \text{y}\text{e}^{2\text{x}}=2\text{x}\ \text{e}^{2\text{x}}-4\times\frac{1}4\text{e}^{2\text{x}}+\text{C}$
$\Rightarrow\ \text{y}\text{e}^{2\text{x}}=2\text{x}\ \text{e}^{2\text{x}}-\text{e}^{2\text{x}}+\text{C}$
$\Rightarrow\ \text{y}\text{e}^{2\text{x}}=2\text{x}\ \text{e}^{2\text{x}}-\text{e}^{2\text{x}}+\text{C}$
$\Rightarrow\ \text{y}\text{e}^{2\text{x}}=(2\text{x}-1)\text{e}^{2\text{x}}+\text{C}$
$\Rightarrow\ \text{y}=(2\text{x}-1)\text{e}^{2\text{x}}+\text{C}\text{e}^{-2\text{x}}$
Hence, $\text{y}=(2\text{x}-1)\text{e}^{2\text{x}}+\text{C}\text{e}^{-2\text{x}}$ is the required solution.
View full question & answer→Question 445 Marks
Solve the following differential equation:$\frac{\text{dy}}{\text{dx}}+2\text{y}=6\text{e}^{\text{x}}$
AnswerWe have,
$\frac{\text{dy}}{\text{dx}}+2\text{y}=6\text{e}^{\text{x}}\ \dots(1)$
Clearly, it is a linear differential equation of the form
$\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
where
$\text{P}=2$
$\text{Q}=6\text{e}^{\text{x}}$
$\therefore$ I.F. $=\text{e}^{\int\text{Pdx}}$
$=\text{e}^{\int2\text{dx}}$
$=\text{e}^{2\text{x}}$
Multiplying both sides of (1) by $\text{e}^{2\text{x}},$ we get
$\text{e}^{2\text{x}}\Big(\frac{\text{dy}}{\text{dx}}+2\text{y}\Big)=6\text{e}^{2\text{x}}\text{e}^\text{x}$
$\Rightarrow\ \text{e}^{2\text{x}}\frac{\text{dy}}{\text{dx}}+2\text{e}^{2\text{x}}\text{y}=6\text{e}^{3\text{x}}$
Integrating both sides with respect to x, we get
$\text{y}\text{e}^{2\text{x}}=6\int\text{e}^{3\text{x}}\text{dx + C}$
$\Rightarrow\ \text{y}\text{e}^{2\text{x}}=6\frac{\text{e}^{3\text{x}}}3+\text{C}$
$\Rightarrow\ \text{ye}^{2\text{x}}=2\text{e}^{3\text{x}}+\text{C}$
Hence, $\text{ye}^{2\text{x}}=2\text{e}^{3\text{x}}+\text{C}$ is the required solution.
View full question & answer→Question 455 Marks
Solve the following differential equations:$\frac{\text{dy}}{\text{dx}}=\frac{\text{x}(2\log\text{x}+1)}{\sin\text{y + y}\cos\text{y}}$
Answerwe have, $\frac{\text{dy}}{\text{dx}}=\frac{\text{x}(2\log\text{x}+1)}{\sin\text{y + y}\cos\text{y}}$
$\Rightarrow(\sin\text{y+y}\cos\text{y})\text{dy = x}(2\log\text{x}+1)\text{dx}$
Integrating both sides, we get
$\int(\sin\text{y+y}\cos\text{y})\text{dy}=\int\text{x}(2\log\text{x}+1)\text{dx}$
$\Rightarrow\int\sin\text{y dy}+\int\text{y}\cos\text{y dy }=2\int\text{x}\log\text{x dx}+\int\text{x dx}$
$\Rightarrow-\cos\text{y}+\Big[\text{y}\int\cos\text{y dy}-\int\Big\{\frac{\text{d}}{\text{dy}}\text{(y)}\int\cos\text{y dy}\Big\}\text{dy}\Big]\\=2\Big[\log\text{x}\int\text{x dx}-\int\Big\{\frac{\text{d}}{\text{dx}}(\log\text{x})\int\text{x dx}\Big\}\text{dx}\Big]+\frac{\text{x}^2}{2}$
$\Rightarrow-\cos\text{y}+\Big[\text{y}\sin\text{y}-\int\sin\text{y dy}\Big]=2\Big[\log\text{x}\times\frac{\text{x}^2}{2}-\int\frac{1}{\text{x}}\times\frac{\text{x}^2}{2}\Big]+\frac{\text{x}^2}{2}$
$\Rightarrow-\cos\text{y+y}\sin\text{y}+\cos\text{y}=\text{x}^2\log\text{x}-\frac{\text{x}^2}{2}+\frac{\text{x}^2}{2}+\text{C}$
$\Rightarrow\text{y}\sin\text{y}=\text{x}^2\log\text{x + C}$
Hence, $\text{y}\sin\text{y = x}^2\log\text{x + C}$ is the required solution.
View full question & answer→Question 465 Marks
Solve the following differential equations:$\tan\text{y}\frac{\text{dy}}{\text{dx}}=\sin(\text{x}+\text{y})+\sin(\text{x}-\text{y})$
Answer$\tan\text{y}\frac{\text{dy}}{\text{dx}}=\sin(\text{x}+\text{y})+\sin(\text{x}-\text{y})$
$\tan\text{y}\frac{\text{dy}}{\text{dx}}=2\sin\Big\{\frac{(\text{x + y})+(\text{x}-\text{y})}{2}\Big\}\cos\Big\{\frac{(\text{x + y})-(\text{x}-\text{y})}{2}\Big\}$
$=2\sin\Big(\frac{\text{x + y + x}-\text{y}}{2}\Big)\cos\Big(\frac{\text{x + y}-\text{ x}+\text{y}}{2}\Big)$
$\tan\text{y}\frac{\text{dy}}{\text{dx}}=2\sin\text{x}\cos\text{y}$
$\frac{\tan\text{y}}{\cos\text{y}}\text{dy}=2\sin\text{x dx}$
$\int\sec\text{y}\tan\text{y dy}=2\int\sin\text{x dx}$
$\sec\text{y}=-2\cos\text{x + C}$
$\sec\text{y}+2\cos\text{x = C}$
View full question & answer→Question 475 Marks
Find the equation of a curve passing through the point (0, 0) and whose differential equation is $\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}}\sin\text{x.}$
AnswerThe differential equation of the curve is:
$\text{y}'=\text{e}^{\text{x}}\sin\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}}\sin\text{x}$
$\Rightarrow\text{dy}=\text{e}^{\text{x}}\sin\text{x}$
Integrating both sides, we get:
$\int\text{dy}=\int\text{e}^{\text{x}}\sin\text{x dx }...(1)$
Let $\text{I}=\int\text{e}^{\text{x}}\sin\text{x dx}.$
$\Rightarrow\text{I}=\sin\text{x}\int\text{e}^{\text{x}}\text{dx}-\int\Big(\frac{\text{d}}{\text{dx}}(\sin\text{x}).\int\text{e}^{\text{x}}\text{dx}\Big)\text{dx}$
$\Rightarrow\text{I}=\sin\text{x}\cdot\text{e}^{\text{x}}-\int\cos\text{x}\cdot\text{e}^{\text{x}}\text{dx}$
$\Rightarrow\text{I}=\sin\text{x}\cdot\text{e}^{\text{x}}-\Big[\cos\text{x}\cdot\int\text{e}^{\text{x}}\text{dx}-\int\Big(\frac{\text{d}}{\text{dx}}(\cos\text{x})\cdot\int\text{e}^{\text{x}}\text{dx}\Big)\text{dx}\Big]$
$\Rightarrow\text{I}=\sin\text{x}\cdot\text{e}^{\text{x}}-\big[\cos\text{x}\cdot\text{e}^{\text{x}}-\int(-\sin\text{x})\cdot\text{e}^{\text{x}}\text{dx}\big]$
$\Rightarrow\text{I}=\text{e}^{\text{x}}\sin\text{x}-\text{e}^{\text{x}}\cos\text{x}-\text{I}$
$\Rightarrow2\text{I}=\text{e}^{\text{x}}(\sin\text{x}-\cos\text{x})$
$\Rightarrow\text{I}=\frac{\text{e}^{\text{x}}(\sin\text{x}-\cos\text{x})}{2}$
View full question & answer→Question 485 Marks
If y(x) is a solution of the different equation $\Big(\frac{2+\sin\text{x}}{1+\text{y}}\Big)\frac{\text{dy}}{\text{dx}}=-\cos\text{x}$ and $\text{y}(0)=1,$ then find the value of $\text{y}\Big(\frac{\pi}{2}\Big).$
AnswerConsider the given equation
$\Big(\frac{2+\sin\text{x}}{1+\text{y}}\Big)\frac{\text{dy}}{\text{dx}}=-\cos\text{x}$
$\Rightarrow\frac{\text{dy}}{(1+\text{y})}=\frac{-\cos\text{x dx}}{(2+\sin\text{x})}$
Integrating both the sides,
$\Rightarrow\int\frac{\text{dy}}{(1+\text{y})}=\int\frac{-\cos\text{x dx}}{(2+\sin\text{x})}$
$\Rightarrow\log(1+\text{y})=-\log(2+\sin\text{x})+\log\text{C}$
$\Rightarrow\log(1+\text{y})+\log(2+\sin\text{x})=\log\text{C}$
$\Rightarrow\log(1+\text{y})(2+\sin)\text{x}=\log\text{C}$
$\Rightarrow(1+\text{y})(2+\sin\text{x})=\text{C}...(1)$
Given that $\text{y}(0)=1$
$\Rightarrow(1+1)(2+\sin0)=\text{C}$
$\Rightarrow\text{C}=4$
Substituting the value of C in equation (1) we have,
$\Rightarrow(1+\text{y})(2+\sin\text{x})=4$
$\Rightarrow(1+\text{y})=\frac{4}{(2+\sin\text{x})}$
$\Rightarrow\text{y}=\frac{4}{(2+\sin\text{x})}-1...(2)$
We need to find the value of $\text{y}\Big(\frac{\pi}{2}\Big)$
Substituting the value of $\text{x}=\frac{\pi}{2}$ in equation (2), we get,
$\text{y}=\frac{4}{\Big(2+\sin\frac{\pi}{2}\Big)}-1$
$\Rightarrow\text{y}=\frac{4}{(2+1)}-1$
$\Rightarrow\text{y}=\frac{4}{3}-1$
$\Rightarrow\text{y}=\frac{1}{3}$
Note: Answer given in the book is incorrect.
View full question & answer→Question 495 Marks
Find the particular solution of $\text{e}^{\frac{\text{dy}}{\text{dx}}}=\text{x}+1,$ that $\text{y}=3,$ when $\text{x}=0.$
Answer$\text{e}^{\frac{\text{dy}}{\text{dx}}}=\text{x}+1$
$\frac{\text{dy}}{\text{dx}}=\log(\text{x}+1),\text{y}=3$ at $\text{x}=0$
$\int\text{dy}=\int\log(\text{x}+1)\text{dx}$
$\text{y}=\log|\text{x}+1|\times\int1\times\text{dx}-\int\Big(\frac{1}{\text{x}+1}\times\int1\text{dx}\Big)\text{dx}+\text{C}$
Using integration by parts
$\text{y = x}\log|\text{x}+1|-\int\frac{\text{x}}{\text{x}+1}\text{dx}+\text{C}$
$\text{y = x}\log|\text{x}+1|-\Big(\int\Big(1-\frac{1}{\text{x}+1}\Big)\text{dx}\Big)+\text{C}$
$=\text{x}\log|\text{x}+1|-(\text{x}-\log|\text{x}+1|)+\text{C}$
$\text{y = x}\log|\text{x}+1|-\text{x}+\log|\text{x}+1|+\text{C}$
$\text{y}=(\text{x}+1)\log|\text{x}+1|-\text{x + C}$
Put $\text{y}=3$ and $\text{x}=0$
$3=0-0+\text{C}$
$\text{C}=3$
Put $\text{C}=3$ in equation (1),
$\text{y}=(\text{x}+1)\log|\text{x}+1|-\text{x}+3$
View full question & answer→Question 505 Marks
Find one-parameter families of solution curves of the following differential equation: (or solve the following differential equation)$\text{e}^{-\text{y}}\sec^2\text{y dy}=\text{dx}+\text{x dy}$
AnswerWe have,
$\text{e}^{-\text{y}}\sec^2\text{y dy}=\text{dx}+\text{x dy}$
$\Rightarrow\text{dx}=\text{e}^{-\text{y}}\sec^2\text{y dy}-\text{x dy}$
$\Rightarrow\frac{\text{dx}}{\text{dy}}=\text{e}^{-\text{y}}\sec^2\text{y}-\text{x}$
$\Rightarrow\frac{\text{dx}}{\text{dy}}+\text{x}=\text{e}^{-\text{y}}\sec^2\text{y}\ ...(1)$
Clearly, it is a linear differential equation of tyhe form
$\frac{\text{dx}}{\text{dy}}+\text{Px}=\text{Q}$
Where
$\text{P}=1$
$\text{Q}=\text{e}^{-\text{y}}\sec^2\text{y}$
$\therefore\ \text{I.F.}=\text{e}^{\int\text{Pdy}}$
$=\text{e}^{\int\text{dy}}$
$=\text{e}^{\text{y}}$
Multiplying both sides of (1) by ey, we get
$\text{e}^{\text{y}}\Big(\frac{\text{dx}}{\text{dy}}+\text{x}\Big)=\text{e}^{\text{y}}\text{e}^{-\text{y}}\sec^2\text{y}$
$\Rightarrow\text{e}^{\text{y}}\frac{\text{dx}}{\text{dy}}+\text{e}^{\text{y}}\text{x}=\sec^2\text{y}$
Integrating both sides with respect to y, we get
$\text{e}^{\text{y}}\text{x}=\int\sec^2\text{y dy}+\text{C}$
$\Rightarrow\text{e}^{\text{y}}\text{x}=\tan\text{y}+\text{C}$
$\Rightarrow\text{x}=(\tan\text{y}+\text{C})\text{e}^{-\text{y}}$
Hence, $\text{x}=(\tan\text{y}+\text{C})\text{e}^{-\text{y}}$ is the required solution.
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