Question 13 Marks
Solve the following differential equation:$4\frac{\text{dy}}{\text{dx}}+\text{8y}=\text{5e}^{-3x}$.
Answer$\frac{\text{dy}}{\text{dy}}+\text{2y}=\frac{5}{4}\text{e}^{-3x}\Rightarrow\text{P(x) = 2, Q(x)}=\frac{5}{4}\text{e}^{-3x}$
I.F. = $\text{e}^{\int\text{p(x)dx}}=\text{e}^{\text{2x}}$
$\therefore \text{We have y}\cdot\text{e}^{\text{2x}}=\int\frac{5}{4}\text{e}^{\text{-3x}}\cdot\text{ e}^{\text{2x}}\text{dx}$
$\Rightarrow\text{y}\cdot\text{e}^{\text{2x}}=\frac{-5}{4}\text{e}^{\text{-x}}+\text{c }\text{ OR }\text{y}=\frac{-5}{4}\text{e}^{\text{-3x}}+\text{c }\text{ e}^{\text{-2x}}.$
View full question & answer→Question 23 Marks
Solve the following differential equation: $y(1 - x^2) \frac{\text{dy}}{\text{dx}} = x(1 + y^2).$
AnswerWriting $y(1 - x^2)$
$ \frac{\text{dy}}{\text{dx}} = x(1 + y^2)$ as
$\int\frac{\text{ydy}}{\text{1 + y}^{2}}=\int\frac{\text{x}}{\text{1 - x}^{2}}\text{dx}$
$\Rightarrow \log|1 + y^2| = - \log|1 - x^2| + \log C_1$
$\Rightarrow (1 + y^2)^{_{. }}(1 - x^2) = C.$
View full question & answer→Question 33 Marks
Form the differential equation of the family of curves y = a cos(x + b), where a and b are arbitrary constants.
Answery = a cos (x + b) $\Rightarrow\frac{\text{dy}}{\text{dx}}$ = - a sin (x + b)$\therefore\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}$ = - a cos (x + b)
$\Rightarrow\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}=-\text{y OR }\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}+\text{y}=0$.
View full question & answer→Question 43 Marks
Solve the following differential equation:
$(y^2 - x^2) dy = 3xy\ dx.$
AnswerWriting $\frac{\text{dy}}{\text{dx}}=\frac{\text{3xy}}{\text{y}^{2}-\text{x}^{2}}=\frac{\text{3y/x}}{\text{y}^{2}/\text{x}^{2}-1}$
Putting $\frac{\text{y}}{\text{x}}=\text{v }\Rightarrow\text{v + x }\frac{\text{dv}}{\text{dx}}=\frac{\text{3v}}{\text{v}^{2}-1}\Rightarrow\text{ x }\frac{\text{dv}}{\text{dx}}=-\frac{\text{v}^{3}-\text{4v}}{\text{v}^{2}-1}$
$\therefore \int\frac{\text{v}^{2}-1}{\text{v}^{3}-\text{4v}}\text{ dv}=-\int\frac{\text{dx}}{\text{x}}\Rightarrow\frac{1}{8}\int\Bigg(\frac{2}{\text{v}}+\frac{3}{\text{v - 2}}+\frac{3}{\text{v + 2}}\Bigg)\text{dv}=-\int\frac{\text{dx}}{\text{x}}$
$\therefore 2 \log v + 3 \log (v - 2) + 3 \log (v + 2) + 8 \log x = \log c$
$\Rightarrow v^2 (v^2 - 4)^3 x^8 = c $
$\Rightarrow y^2 (y^2 - 4x^2)^3 = c.$
View full question & answer→Question 53 Marks
Solve the following differential equation:
$\sin x \frac{dy}{dx} + \cos x y = \cos x \sin ^{2} x$
AnswerGiven equation can be written as
$\frac{dy}{dx} + \cot x . y- \cos x . \sin x \therefore \text{Integrating factor} = e^{\log\sin x} = \text{\sin x} $
$\therefore \text{y} . \sin \text{x} = \int \sin ^{2}\text{ x.} \cos \text{x. dx}$
$\Rightarrow \text{y}. \sin\text{x} = \frac{\sin^{3}\text{x}}{3} \text{+ c or } \text{y} = \frac{1}{3} \sin^{2}\text{x + c}. \text{cosec x}$
View full question & answer→Question 63 Marks
Verify that y = A cos x - b sin x is a solution of the differential equation.$\frac{\text{d}^{2} \text{y}}{\text{dx}^{2}}+\text{y}=0.$
Answer$\frac{\text{dy}}{\text{dx}}=-\text{A sin x - B cos x}$
$\therefore\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}} =-\text{A cos x + B sin x}$
$\therefore\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}=-\text{(A cos x - B sin x})=-\text{y }\text{ }\therefore\text{ }\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}+\text{y}=0$
$\therefore$ y = A cos x - B sin x is the solution of given diffential equation.
View full question & answer→Question 73 Marks
Form the differential equation of the family of curves $y = A\cos 2 x + B\sin2x,$where A and B are constants.
Answer$y = A\cos 2 x + B\sin2x,$$\therefore \frac{dy}{dx} = -\text{2 A} \sin 2x + \text{2 B}\cos 2x$
$\frac{d^{2} y}{dx^{2}} = -\text{4 A} \cos 2x - \text{4 B}\sin 2x$
$\frac{d^{2} y}{dx^{2}} = -\text{4 . y}$
or $\frac{d^{2}y}{dx^{2}} + 4y = 0$ is the required differential equation.
View full question & answer→Question 83 Marks
Solve the following differential equation:$\frac{dy}{dx} + 2y = \text6 {e^{x}}$
Answer$\text{I.F.} = e^{\int 2dx} = e^{2x}$$\therefore \text{the solution is} $
$y . e^{2{x}} = 6 \int e^{3x} \text{dx + c} $
$y . e^{2x} ={\text{2 e}}^{3x}+ c$
or $y = \text{2 e}^{x} + {\text{c e}^{-2x}}$
View full question & answer→Question 93 Marks
Solve the following differential equation:$2 xydx + (x^{2} +2y^2) dy = 0$
Answer$\frac{dy}{dx} = \frac{- 2xy}{x^{2}+ 2y^{2}} ; \text{Let} y = vx\Rightarrow \frac{dy}{dx} = v + x \frac{dy}{dx}$$\therefore v + x \frac{dv}{dx} = \frac{- 2vx^2}{x + 2v^2x^2} = \frac{- 2v}{1 + 2v^2}$
$\therefore x \frac{dv}{dx} = \frac{- 2v - v - 2v^{3}}{1 + 2v^{2}} = -\frac{2v^{3} + 3v}{1 + 2v^{2}}$
$\therefore \frac{dx}{x} = \frac{1 + 2v^{2}}{(2v^{2} + 3)v}$
$= -\bigg[ \frac{2v^{2} + 3 - 2}{v (2v^{2} +3}\bigg] dv = \frac{-dv}{v} + \frac{2dv}{v (2v^{2} + 3)}$
$\therefore \frac{dx}{v} = \frac{-dv}{v} + \bigg[ \frac{A}{v} + \frac{Bv + C}{2v^{2} +3}\bigg]$
$A = \frac{2}{3}, B = -\frac{4}{3}, C = 0$
$\therefore \frac{dx}{x} = -\frac{dv}{v} + \frac{2}{3} \frac{dv}{v} -\frac{1}{3} \frac{4vdv}{2v^{2} + 3}$
Intergrating, we get
$\log x = -\log v + \frac{2}{3} \log v -\frac{1}{3} \log(2v^{2} + 3) + \log \text{c}$
$= -\frac{1}{3} \log v -\frac{1}{3} \log (2v^{2} + 3) + \log c$
$\log x^-3 = \log cv(2v^{2} + 3)$ or
$\frac{1}{x^{3}} = c \frac{y}{x} \bigg(\frac{2y^{2}}{x^{2}} + 3 \bigg)$ or
$= cy \bigg(\frac{2y^{2}}{x^{3}} + \frac{3x^{2}}{x^{3}}\bigg)$ or
$cy (2y^{2} + 3x^{2} = 1$
View full question & answer→Question 103 Marks
Form the differential equation of the family of curves y = a sin (x + b), where a and b are arbitrary constants.
Answer$y = \text {a}\sin (x + b)$$\therefore \frac {dy}{dx} = \text {a}\cos (x + b)$
$\frac{d^2y}{dx^2} = -\text {a}\sin (x+ b) =-y$
$\therefore \frac{d^{2}y}{dx^{2}} + y = 0$
View full question & answer→Question 113 Marks
Solve the following differential equation:$\frac{dy}{dx} -\frac{y}{x} = 2x^{2}$
AnswerIntegrating factor $= e^{-\int\frac{dx}{x}}$$= e^{-\int\frac{- logx}{}} = \frac{1}{x} $
The solution of different equation is
$ y^\frac{1} {x} = {\int\frac{2x^{2}}{x}} dx + c$
$y^{{}{}} \frac{1}{x} = x^{2} + c$ or
$y = x^{3} + cx$
View full question & answer→Question 123 Marks
Show that $\text{y}=\text{Ae}^{\text{bx}}$ is a solution of the differential equation $\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{1}{\text{y}}\Big(\frac{\text{dy}}{\text{dx}}\Big)^2.$
AnswerWe have,
$\text{y}=\text{Ae}^{\text{bx}}\ ...(1)$
Differentiating both sides of equation (1) with respect to 3, we get
$\frac{\text{dy}}{\text{dx}}=\text{ABe}^{\text{Bx}}\ ...(2)$
Differentiating both sides of equation (2) with respect to 3, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\text{AB}^2\text{e}^{\text{Bx}}$
$=\frac{(\text{ABe}^{\text{Bx}^2})}{(\text{Ae}^{\text{Bx}})}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{1}{\text{y}}\Big(\frac{\text{dy}}{\text{dx}}\Big)^2$
Hence, the given function is the solution to the given differential equation.
View full question & answer→Question 133 Marks
For each of the differential equations given in find the general solution:
$\frac{\text{dy}}{\text{dx}}+3\text{y}=\text{e}^{-2\text{x}}$
AnswerThe given differential equation is $\frac{\text{dy}}{\text{dx}}+\text{py}=\text{Q} ($where $p = 3$ and $Q = e ^{-2x})$
$\text{Now, I.F}=\text{e}^{\int\text{pdx}}=\text{e}^{\int3\text{dx}}=\text{e}^{3\text{x}}.$
The solution of the given differential equation is given by the relation,
$\text{y}(\text{I.F})=\int(\text{Q}\times\text{I.F.})\text{dx}+\text{C}$
$\Rightarrow\ \text{ye}^{\text{3}\text{x}}=\int(\text{e}^{-2\text{x}}\times\text{e}^{3\text{x}})+\text{C}$
$\Rightarrow\ \text{ye}^{\text{3}\text{x}}=\int\text{e}^\text{x}\text{dx}+\text{C}$
$\Rightarrow\ \text{ye}^{\text{3}\text{x}}=\text{e}^\text{x}+\text{C}$
$\Rightarrow\ \text{y}=\text{e}^{\text{-2}\text{x}}+\text{C e}^{\text{-3}\text{x}}$
This is the required general solution of the given differential equation.
View full question & answer→Question 143 Marks
For each of the differential equations given in find the general solution:
$\frac{\text{dy}}{\text{dx}}+\sec\text{xy}=\tan\text{x}\Big(0\leq\text{x}<\frac{\pi}{2}\Big)$
AnswerThe given differential equation is:$\frac{\text{dy}}{\text{dx}}+\text{py}=\text{Q (where p}=\sec\text{x and Q}=\tan\text{x})$
$\text{Now, I.F}=\text{e}^{\int\text{pdx}}=\text{e}^{\int\sec\text{x dx}}=\text{e}^{\log(\sec\text{x}+\tan\text{x})}=\sec.\text{x}+\tan\text{x}.$
The general solution of the given differential equation is given by the relation,
$\text{y}(\text{I.F})=\int(\text{Q}\times\text{I.F.})\text{dx}+\text{C}$
$\Rightarrow\ \text{y}(\sec\text{x}+\tan\text{x})=\int\tan\text{x}(\sec\text{x}+\tan\text{x})\text{dx}+\text{C}$
$\Rightarrow\ \text{y}(\sec\text{x}+\tan\text{x})=\int\sec\text{x}\tan\text{x}\text{dx}+\int\tan^2\text{x}\text{dx}+\text{C}$
$\Rightarrow\ \text{y}(\sec\text{x}+\tan\text{x})=\sec\text{x}+\int(\sec^2 \text{x}-1)\text{dx}+\text{C}$
$\Rightarrow\ \text{y}(\sec\text{x}+\tan\text{x})=\sec\text{x}+\tan\text{x}-\text{x}+\text{C}$
This is the required general solution of the given differential equation.
View full question & answer→Question 153 Marks
Represent the following families of curves by forming the corresponding differential equation:
$\text{y}^2=4\text{a}(\text{x}-\text{b})$
AnswerThe equation of the family of curves is
$\text{y}^2=4\text{a}(\text{x}-\text{b})\ ...(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{y}\frac{\text{dy}}{\text{dx}}=4\text{a}$
$\Rightarrow\text{y}\frac{\text{dy}}{\text{dx}}=2\text{a}$
Differentiating (2) with respect to x, we get
$\text{y}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2=0$
It is the required differential equation.
View full question & answer→Question 163 Marks
For each of the differential equations given in find the general solution:$\text{y dx}+(\text{x}-\text{y}^{2})\ \text{dy}=0$
Answer$\text{y dx}+(\text{x}-\text{y}^{2})\text{dy}=0$$\Rightarrow\frac{\text{dx}}{\text{dy}}=\frac{\text{y}^2-\text{x}}{\text{y}}=\text{y}-\frac{\text{x}}{\text{y}}$
$\Rightarrow\frac{\text{dx}}{\text{dy}}+\frac{\text{x}}{\text{y}}=\text{y}$
This is a linear differential equation of the form:
$\frac{\text{dy}}{\text{dx}}+\text{px}=\text{Q}\ (\text{where p}=\frac{1}{\text{y}\ }\ \text{and}\ \text{Q}=\text{y})$
$\text{Now, I.F}=\text{e}^{\int\text{pdy}}=\text{e}^{\int\frac{1}{\text{y}}\text{dy}}=\text{e}^{\log\text{y}}=\text{y}.$
The general solution of the given differential equation is given by the relation,
$\text{y(I.F)}=\int(\text{Q}\times\text{I.F.})\text{dx}+\text{C}$
$\Rightarrow\text{xy}=\int(\text{y}\cdot\text{y})\text{dy}+\text{C}$
$\Rightarrow\text{xy}=\int\text{y}^2\text{dy}+\text{C}$
$\Rightarrow\text{xy}=\frac{\text{y}^3}{3}+\text{C}$
$\Rightarrow\text{x}=\frac{\text{y}^2}{3}+\frac{\text{C}}{\text{y}}$
View full question & answer→Question 173 Marks
Solve the following differential equations:$\frac{\text{dy}}{\text{dx}}+\frac{\cos\text{x}\sin\text{y}}{\cos\text{y}}=0$
Answer$\frac{\text{dy}}{\text{dx}}+\frac{\cos\text{x}\sin\text{y}}{\cos\text{y}}=0$
$\frac{\text{dy}}{\text{dx}}=-\cos\text{x}\tan\text{y}$
$\frac{\text{dy}}{\tan\text{y}}=-\cos\text{x dx}$
$\int\cot\text{ y dy}=-\int\cos\text{x dx}$
$\log|\sin\text{y}|=-\sin\text{x + C}$
$\sin\text{x}+\log|\sin\text{y}|=\text{C}$
View full question & answer→Question 183 Marks
Solve the following differential equations:$\frac{\text{dr}}{\text{dt}}=-\text{rt, r}(0)=\text{r}_{0}$
Answer$\frac{\text{dr}}{\text{dt}}=-\text{rt, r}(0)=\text{r}_{0}$
$\int\frac{\text{dr}}{\text{r}}=-\int\text{tdt}$
$\log|\text{r}|=-\frac{\text{t}^2}{2}+\text{C}...(1)$
Put $\text{t = 0, r = r}_{0}$ inequation (1),
$\log|\text{r}_0|=0+\text{C}$
$\log|\text{r}_0|=\text{C}$
Now,
$\log|\text{r}|=-\frac{\text{t}^2}{2}+\log|\text{r}_0|$
$\frac{\text{r}}{\text{r}_0}=\text{e}^{-\frac{\text{t}^2}{2}}$
$\text{r}=\text{r}_0\text{e}^{-\frac{\text{t}^2}{2}}$
View full question & answer→Question 193 Marks
Solve the following differential equations:$\text{ye}^{\frac{\text{x}}{\text{y}}}\text{dx}=(\text{xe}^{\frac{\text{x}}{\text{y}}}+\text{y}^2)\text{dy, y}\neq0$
Answer$\text{ye}^{\frac{\text{x}}{\text{y}}}\text{dx}=(\text{xe}^{\frac{\text{x}}{\text{y}}}+\text{y}^2)\text{dy}$
$\Rightarrow\text{ye}^{\frac{\text{x}}{\text{y}}}\text{dx = xe}^{\frac{\text{x}}{\text{y}}}\text{dy}+\text{y}^2\text{dy}$
$\Rightarrow\text{ye}^{\frac{\text{x}}{\text{y}}}\text{dx}-\text{xe}^{\frac{\text{x}}{\text{y}}}\text{dy = y}^2\text{dy}$
$\Rightarrow(\text{ydx}-\text{xdy})\text{e}^{\frac{\text{x}}{\text{y}}}=\text{y}^2\text{dy}$
$\Rightarrow\frac{(\text{ydx}-\text{xdy})}{\text{y}^2}\text{e}^{\frac{\text{x}}{\text{y}}}=\text{dy}$
$\Rightarrow\text{e}^{\frac{\text{x}}{\text{y}}}\text{d}\Big(\frac{\text{x}}{\text{y}}\Big)=\text{dy}$
$\Rightarrow\int\text{e}^{\frac{\text{x}}{\text{y}}}\text{d}\Big(\frac{\text{x}}{\text{y}}\Big)=\int\text{dy}$
$\Rightarrow\text{e}^{\frac{\text{x}}{\text{y}}}=\text{y + C}$
View full question & answer→Question 203 Marks
Solve the following differential equations:$\frac{\text{dy}}{\text{dx}}=\text{y}\sin2\text{x, y}(0)=1$
AnswerWe have,
$\frac{\text{dy}}{\text{dx}}=\text{y}\sin2\text{x, y}(0)=1$
$\Rightarrow\frac{1}{\text{y}}\text{dy}=\sin2\text{x dx}$
Integrating both sides, we get
$\int\frac{1}{\text{y}}\text{dy}=\int\sin2\text{x dx}$
$\Rightarrow\log|\text{y}|=-\frac{\cos 2\text{x}}{2}+\text{C}...(1)$
Given: $\text{x}=0,\text{y}=1.$
Substituting the values of x and y in (1), we get
$\log|1|=-\frac{1}{2}+\text{C}$
$\Rightarrow\text{C}=\frac{1}{2}$
Substituting the values of C in (1), we get
$\log|\text{y}|=-\frac{\cos 2\text{x}}{2}+\frac{1}{2}$
$\Rightarrow\log|\text{y}|=\frac{1-\cos 2\text{x}}{2}$
$\Rightarrow\log|\text{y}|=\sin^2\text{x}$
$\Rightarrow\text{y}=\text{e}^{\sin^2\text{x}}$
Hence, $\text{y}=\text{e}^{\sin^2\text{x}}$ is the required solution
View full question & answer→Question 213 Marks
Differential equation $\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{y}=0,\text{y}(0)=0,\text{y}'(0)=1$Function $\text{y}=\sin\text{x}$
AnswerWe have,$\text{y}=\sin{\text{x}} ...(1)$
Differentiating both sides of (1) with respect to x, we get
$\frac{\text{dy}}{\text{dx}}=\cos\text{x} ...(2)$
Differentiating both sides of (2) with respect to x, we get
$\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}=-\sin{\text{x}}$
$\Rightarrow\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}=-\text{y}$ [Using (1)]
$\Rightarrow \frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}+\text{y}=0$
it is the given differential equation.
Here, $\text{y}=\sin\text{x}$ satisfies the given differential equation; hence, it is a solution.
Also, when $\text{x}=0,\text{y}=\sin0=0,\text{i.e.},\text{y}(0)=0.$
And, when $\text{x}=0,\text{y}'=\cos 0=1,\text{i.e.,}\text{y}'(0)=1.$
Hence, $\text{y}=\sin\text{x}$ is the solution to the given initial value problem.
View full question & answer→Question 223 Marks
Solve the following differential equations:$(1-\text{x}^2)\text{dy + xy dx = xy}^2\text{ dx}$
Answer$(1-\text{x}^2)\text{dy + xy dx = xy}^2\text{dx}$
$(1-\text{x}^2)\text{dy = dx}(\text{xy}^2-\text{xy})$
$(1-\text{x}^2)\text{dy = xy(y}-1)\text{dx}$
$\int\frac{\text{dy}}{\text{y(y}-1)}=\int\frac{\text{xdx}}{1-\text{x}^2}$
$\int\Big(\frac{1}{\text{y}-1}-\frac{1}{\text{y}}\Big)\text{dy}=\frac{1}{2}\int\frac{2\text{x}}{1-\text{x}^2}\text{dx}$
$\int\Big(\frac{1}{\text{y}-1}-\frac{1}{\text{y}}\Big)\text{dy}=-\frac{1}{2}\int\frac{-2\text{x}}{1-\text{x}^2}\text{dx}$
$\log|\text{y}-1|-\log|\text{y}|=-\frac{1}{2}\log|1-\text{x}^2|+\text{C}$
View full question & answer→Question 233 Marks
Differential equation $\frac{\text{dy}}{\text{dx}}+\text{y}=0,\text{y}(0)=3$ Function $\text{y}=\text{e}^\text{-x}+2$
AnswerHere, $\text{y}=\text{e}^{\text{x}}+1 ....(1)$
Differentiating it with respect to $x,$
$\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}}$
$\frac{\text{dy}}{\text{dx}}=\text{y}-1 ...(2)$
Again, differentiating it with respect to $x,$
$\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}=\frac{\text{dy}}{\text{dx}}$
$\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}-\frac{\text{dy}}{\text{dx}}=0$
It is given differential equation.
so,$y = e^{x }+ 1$ is a solution of the equation
put $x - 0$ in equation $(1),$
$\Rightarrow y = e^{0 }+ 1 = 2$
$y(0) = 2$
put $x = 0$ in equation $(2),$
$y' = e^0 = 1$
$y(0) = 1$
View full question & answer→Question 243 Marks
Differential equation $\text{x}\frac{\text{dy}}{\text{dx}}=1,\text{y}(1)=0$Function $\text{y}=\log\text{x}$
AnswerHere, y = logxDifferentiating it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}}$
$\text{x}\frac{\text{dy}}{\text{dx}}=1$
so, y=logx is a solution of the equation
If $\text{x}=1,\text{y}=\log1=0$
so,
y(1) = 0
View full question & answer→Question 253 Marks
For each of the differential equations in find the general solution:
$\frac{\text{dy}}{\text{dx}}=(1+\text{x}^2)(1+\text{y}^2)$
AnswerGiven: Differential equation $\frac{\text{dy}}{\text{dx}}=(1+\text{x}^2)(1+\text{y}^2)$
$\Rightarrow\ \text{dy}=\big(1+\text{x}^2\big) \big(1+\text{y}^2\big)\text{dx}\ \Rightarrow \ \frac{\text{dy}}{1+\text{y}^2}$ $=\big(1+\text{x}^2\big)\text{dx}\ \ [\text{Separating variables}]$
Integrating both sides, $\int \frac{1}{\text{y}^2+1}\text{dy}=\int\big(\text{x}^2+1\big)\text{dx}$
$\Rightarrow \ \ \ \ \tan^{-1}\text{y}=\frac{\text{x}^3}{3}+\text{x+c}$
View full question & answer→Question 263 Marks
In each of the form a differential equation representing the given family of curves by eliminating arbitrary constants $a$ and $b. y^2 = a (b^2 – x^2)$
Answer$y^2 = a (b^2 – x^2)$
Differentiating both sides with respect to $x,$ we get:
$2\text{y}\frac{\text{dy}}{\text{dx}}=a(-2\text{x})$
$\Rightarrow 2\text{yy}\ '=-2\text{ax}$
$\Rightarrow \text{yy}\ '=-\text{ax}\ \ ...(1)$
Again, differentiating both sides with respect to $x,$ we get:
$\text{y}\ '.\text{y}\ '+\text{yy}\ ''=-\text{a}$
$\Rightarrow (\text{y}\ ')^2+\text{yy}\ ''=-\text{a} \ \ ...(2)$
Dividing equation $(2)$ by equation $(1),$ we get:
$\frac{(\text{y}\ ')^2+\text{yy}\ ''}{\text{yy}\ '}=\frac{-\text{a}}{-\text{ax}}$
$\Rightarrow \text{xyy}\ ''+\text{x(y}\ ')^2-4\text{y}\ ' = 0$
This is the required differential equation of the given curve.
View full question & answer→Question 273 Marks
For each of the differential equations given in find the general solution:
$(\text{x}+3\text{y}^2)\frac{\text{dy}}{\text{dx}}=\text{y}(\text{y}>0)$
Answer$(\text{x}+3\text{y}^2)\frac{\text{dy}}{\text{dx}}=\text{y}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}+3\text{y}^2}$
$\Rightarrow\frac{\text{dx}}{\text{dy}}=\frac{\text{x}+3\text{y}^2}{\text{y}}=\frac{\text{x}}{\text{y}}+3\text{y}$
$\Rightarrow\frac{\text{dx}}{\text{dy}}-\frac{\text{x}}{\text{y}}=3\text{y}$
This is a linear differential equation of the form:
$\frac{\text{dx}}{\text{dy}}+\text{px}=\text{Q}\ (\text{where p}=-\frac{1}{\text{y}}\ \text{and}\ \text{Q}=3\text{y})$
$\text{Now, I.F}=\text{e}^{\int\text{pdy}}=\text{e}^{-\int\frac{\text{dy}}{\text{y}}}=\text{e}^{-\log\text{y}}=\text{e}^{\log\Big(\frac{1}{\text{y}}\Big)}=\frac{1}{\text{y}}.$
The general solution of the given differential equation is given by the relation,
$\text{x}\text{(I.F.)}=\int(\text{Q}\times\text{I.F.})\text{dy}+\text{C}$
$\Rightarrow\text{x}\times\frac{1}{\text{y}}=\int\Big(3\text{y}\times\frac{1}{\text{y}}\Big)\text{dy}+\text{C}$
$\Rightarrow\frac{\text{x}}{\text{y}}=3\text{y}+\text{C}$
$\Rightarrow\text{x}=3\text{y}^2+\text{Cy}$
View full question & answer→Question 283 Marks
For each of the differential equations in Exercises find the general solution:
$e^x \tan y dx + (1 – e^x) \sec^2\ y dy = 0$
AnswerGiven: Differential equation $e^x \tan y dx + (1 – e^x) \sec^2\ y dy = 0$
Dividing each term by $(1-e^x) \tan y,$ we get
$\frac{\text{e}^\text{x}}{1-\text{e}^\text{x}} \text{dx}+\frac{\text{sec}^2\text{y}}{\tan \text{y}} \text{dy = 0}[$Separating variables$]$
Integrating both sides, $\ \int \frac{\text{e}^\text{x}}{1-\text{e}^\text{x}} \text{dx}+\int\frac{\text{sec}^2\text{y}}{\tan \text{y}} \text{dy}=\text{c}$
$\Rightarrow \ -\int\frac{-\text{e}^\text{x}}{1-\text{e}^\text{x}}\text{dx + log|tan y| = c} $
$ \Rightarrow \ -\text{log|1-}\text{e}^\text{x}|+\text{log}|\text{tan y}|=\text{c}$
$\Rightarrow \ \text{log}\frac{|\tan \text{y|}}{|1-\text{e}^\text{x}|}=\text{log c}\ '$
$\Rightarrow \ \frac{|\tan \text{y|}}{|1-\text{e}^\text{x}|} = \text{c}\ '$
$\Rightarrow \ \tan \text{y} = \text{C}\big(1-\text{e}^\text{x}\big) \ \big[\because|\text{t}|=\text{c}\ ' =\text{C (say)}\big]$
View full question & answer→Question 293 Marks
Solve $\text{y}\text{dx}-\text{x}\text{dy}=\text{x}^2\text{y}\text{dx}.$
AnswerGiven that, $\text{y}\text{dx}-\text{x}\text{dy}=\text{x}^2\text{y}\text{dx}$
$\Rightarrow\frac{1}{\text{x}^2}-\frac{1}{\text{xy}}.\frac{\text{dy}}{\text{dx}}=1$ [dividing throughout by $x^2ydx]$
$\Rightarrow-\frac{1}{\text{xy}}.\frac{\text{dy}}{\text{dx}}+\frac{1}{\text{x}^2}-1=0$
$\Rightarrow\frac{\text{dy}}{\text{dx}}-\frac{\text{xy}}{\text{x}^2}+\text{xy}=0$
$\Rightarrow\frac{\text{dy}}{\text{dx}}-\frac{\text{y}}{\text{x}}+\text{xy}=0$
$\Rightarrow\frac{\text{dy}}{\text{dx}}+\Big(\text{x}-\frac{1}{\text{x}}\Big)\text{y}=0$
Which is linear differential equation.
On comparing it with $\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q},$ we get
$\text{P}=\Big(\text{x}-\frac{1}{\text{x}}\Big),\text{Q}=0$
$\text{I.f}=\text{e}^{\int\text{P}\text{dx}}$
$=\text{e}^{\int\big(\text{x}-\frac{1}{\text{x}}\big)}\text{dx}$
$=\text{e}^{\frac{\text{x}^2}{2}-\log\text{x}}$
$=\text{e}^{\frac{\text{x}^2}{2}}.\text{e}^{-\log\text{x}}$
$=\frac{1}{\text{x}}\text{e}^{\frac{\text{x}^2}{2}}$
The general solution is,
$\text{y}.\frac{1}{\text{x}}\text{e}^\frac{\text{x}^2}{2}=\int0.\frac{1}{\text{x}}\text{e}^{\frac{\text{x}^2}{2}}\text{dx}+\text{C}$
$\Rightarrow\text{y}.\frac{1}{\text{x}}\text{e}^{\frac{\text{x}^2}{2}}=\text{C}$
$\Rightarrow\text{y}=\text{C}\text{x}\text{e}^{\frac{-\text{x}^2}{2}}$
View full question & answer→Question 303 Marks
Find the equation of a curve passing through the point $(0, 0)$ and whose differential equation is $y\ ' = e^x \sin x.$
AnswerThe given differential equation is $\text{y ' }=\text{e}^{\text{x}}\ \text{sin}\ \text{x} $ or $\ \frac{\text{dy}}{\text{dx}}= \text{e}^{\text{x}}\ \text{sin}\ \text{x}$
Separting the variables, we get $,dy = e^x \sin \ x dx$
Integrating, $\int\ \text{dy}=\int\text{e}^\text{x}\ \text{sin x}\ \text{ dx}$
$\therefore\ \text{y}=\frac{1}{1+1}\text{e}^\text{x}(\text{sin x - cos x})+\text{c}$
$\Bigg[\because\ \int\text{e}^\text{ax}\ \text{sin b x dx}=\frac{1}{\text{a}^2+\text{b}^2}\text{e}^\text{ax}(\text{a sin bx - b cos bx})\Bigg]$
$\therefore\ \text{y}=\frac{1}{2}\text{e}^\text{x}(\text{sin x - cos x})+\text{c}\ ....(1)$
Now curve passes through $(0, 0)$
$\therefore\ 0=\frac{1}{2}\text{e}^0(\text{sin}\ 0-\text{cos}\ 0)+\text{c}$
$\Rightarrow\ 0=\frac{1}{2}\text{x}(0 - 1)+\text{c}$
$\therefore\ \text{c}=\frac{1}{2}$
$\therefore$ from $(1),\ \text{y}=\frac{1}{2}\text{e}^{\text{x}}(\text{sin x - cos x})+\frac{1}{2}$ which is required solution.
View full question & answer→Question 313 Marks
For each of the exercises given below, verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.
| $\text{x}^{2}=2\text{y}^2\log\text{y}$ |
: |
$(\text{x}^2+\text{y}^2)\frac{\text{dy}}{\text{dx}}-\text{xy}=0$ |
AnswerThe given differential equation is
$\text{x}^2=2\text{y}^2\log\text{y}\ \ ...(1)$
$\therefore\ \ 2\text{x}=2\Big\{\text{y}^2\times\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}+\log\text{y}\Big[2\text{y}\frac{\text{dy}}{\text{dx}}\Big]\Big\}$
$\text{or}\ \ 2\text{x}=2(\text{y}+2\text{y}\log\text{y})\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\ \ \text{x}=(\text{y}+2\text{y}\log\text{y})\frac{\text{dy}}{\text{dx}}$
Multiplying both sides by y, we get
$\text{xy}=(\text{y}^2+2\text{y}^2\log\text{y})\frac{\text{dy}}{\text{dx}}$
$\text{or}\ \ \text{xy}=(\text{y}^2+\text{x}^2)\frac{\text{dy}}{\text{dx}}\ \ [\because \text{of}\ (1)]$
$\text{or}\ \ (\text{x}^2+\text{y}^2)\frac{\text{dy}}{\text{dx}}-\text{xy}=0$
Hence the result.
View full question & answer→Question 323 Marks
Find the solution of the differential equation $\text{x}\sqrt{1+\text{y}^{2}}\text{dx}+\text{y}\sqrt{1+\text{x}^{2}}\text{dy}=0.$
Answer$\text{x}\sqrt{1+\text{y}^{2}}\text{dx}+\text{y}\sqrt{1+\text{x}^{2}}\text{dy}=0$
$\Rightarrow \text{y}\sqrt{1+\text{x}^{2}}\text{dy}=-\text{x}\sqrt{1+\text{y}^{2}}\text{dx}=0$
$\Rightarrow \frac{\text{y}}{\sqrt{1+\text{y}^{2}}}\text{dy}=\frac{-\text{x}}{\sqrt{1+\text{x}^{2}}}\text{dx}$
$\Rightarrow \int\frac{\text{y}}{\sqrt{1+\text{y}^{2}}}\text{dy}=-\int\frac{\text{x}}{\sqrt{1+\text{x}^{2}}}\text{dx}$
Let $1+\text{y}^{2}=\text{t}^{2}=\text{t}^{2}, 1+\text{x}^{2}=\text{p}^{2}$
$\Rightarrow 2\text{y}\ \text{dy}=2\text{t}\ \text{dt}, 2\text{x}\ \text{dx}=2\text{p}\ \text{dp}$
$\Rightarrow \text{y}\ \text{dy}=\text{t}\ \text{dt}, \text{x}\ \text{dx}=\text{p}\ \text{dp}$
Substituting in above equation, we get
$\Rightarrow \int\text{dt}=-\int\text{dp}$
$\Rightarrow \text{t}=-\text{p}+\text{C}$
$\Rightarrow\sqrt{1+\text{x}^{2}}+\sqrt{1+\text{y}^{2}}=\text{C}$
View full question & answer→Question 333 Marks
Form the differential equation representing the family of curves given by $(\text{x}-\text{a})^2+2\text{y}^2=\text{a}^2,\ \text{where a}$ is an arbitrery constant.
AnswerEquation of the given family of curves is $(\text{x}-\text{a})^2+2\text{y}^2=\text{a}^2$
$\Rightarrow\ \ \text{x}^2+\text{a}^2-2\text{ax}+2\text{y}^2=\text{a}^2\ \ $ $\Rightarrow\ \ \text{x}^2-2\text{ax}+2\text{y}^2=0$
$\Rightarrow\ \ \text{x}^2+2\text{y}^2=2\text{ax}^\ \ ....\text{(i)}$
Here number of arbitrary constants is one only (a).
So we will differentiate both sides of equation only once w,r.t. x,
$\Rightarrow\ \ 2\text{x}+2.2\text{y}\frac{\text{dy}}{\text{dx}}=2\text{a}$ $\Rightarrow\ \ 2\text{x}+4\text{y}\frac{\text{dy}}{\text{dx}}=2\text{a}\ \ ...\text{(ii)}$
Dividing eq. (i) by eq. (ii), we have $\frac{\text{x}^2+2\text{y}^2}{2\text{x}+4\text{y}\frac{\text{dy}}{\text{dx}}}=\frac{2\text{ax}}{2\text{a}}\ \ \Rightarrow\ \ \frac{\text{x}^2+2\text{y}^2}{2\text{x}+4\text{y}\frac{\text{dy}}{\text{dx}}}=\text{x}$
$\Rightarrow\ \ \text{x}\Big(2\text{x}+4\text{y}\frac{\text{dy}}{\text{dx}}\Big)=\text{x}^2+2\text{y}^2$ $\Rightarrow\ \ 2\text{x}^2+4\text{xy}\frac{\text{dy}}{\text{dx}}=\text{x}^2+2\text{y}^2$
$\Rightarrow\ \ 4\text{xy}\frac{\text{dy}}{\text{dx}}=2\text{y}^2-\text{x}^2\ \ \Rightarrow\ \ \frac{\text{dy}}{\text{dx}}=\frac{2\text{y}^2-\text{x}^2}{4\text{xy}}$
View full question & answer→Question 343 Marks
Find the particular solution of the differential equation $(1+\text{e}^{2\text{x}})\text{dy}+(1+\text{y}^2)\text{e}^{\text{x}}\ \text{dx}=0,\ \text{given that y}=1\ \text{when x}=0. $
AnswerThe given differential equation is
$(1+\text{e}^{2\text{x}})\text{dy}+(1+\text{y}^2)\text{e}^\text{x}\ \text{dx}=0$ $\text{or}\ \ (1+\text{e}^{2\text{x}})\text{dy}=-(1+\text{y}^2)\text{e}^{\text{x}}\ \text{dx}$
$\therefore\ \ \frac{1}{1+\text{y}^2}\text{dy}=-\frac{\text{e}^{\text{x}}}{1+\text{e}^{2\text{x}}}\text{dx}$
$\Rightarrow\ \ \int\frac{1}{1+\text{y}^2}\text{dy}=-\int\frac{\text{e}^\text{x}}{1+\text{e}^{2\text{x}}}\text{dx}\ \ ...(1)$
$\text{Let l}=\int\frac{\text{e}^\text{x}}{1+\text{e}^{2\text{x}}}1+\text{dx}$
$\text{Put}\ \ \text{e}^\text{x}=\text{t},\ \therefore\ \text{e}^\text{x}\ \text{dx}=\text{dt}$
$\therefore\ \text{l}=\int\frac{\text{dt}}{1+\text{t}^2}=\tan^{-1}\text{t}=\tan^{-1}(\text{e}^\text{x})$
$\therefore\ \text{form}\ (2),\ \tan^{-1}\text{y}=\tan(\text{e}^\text{x})+\text{c}\ \ ...(2)$
$\text{Now}, \ \text{x}=0,\ \text{y}=1$
$\therefore\ \tan^{-1}1=-\tan(\text{e}^0)+\text{c}$ $\Rightarrow\ \ \frac{\pi}{4}=-\tan^{-1}(1)+\text{c}$
$\Rightarrow\ \ \frac{\pi}{4}=-\frac{\pi}{4}+\text{c}\ \ \Rightarrow\ \ \text{c}=\frac{\pi}{2}$
$\therefore\ \ \text{form}\ (2), \tan^{-1}=-\tan\text{e}^{\text{x}}+\frac{\pi}{2},$ which is required solution.
View full question & answer→Question 353 Marks
Solve the differential equation $\bigg[\frac{\text{e}^{-2\sqrt{\text{x}}}}{\sqrt{\text{x}}}-\frac{\text{y}}{\sqrt{\text{x}}}\bigg]\frac{\text{dx}}{\text{dy}}=1(\text{x}\neq0).$
AnswerGiven: Differential equation: $\bigg(\frac{\text{e}^{-2\sqrt{\text{x}}}}{\sqrt{\text{x}}}-\frac{\text{y}}{\sqrt{\text{x}}}\bigg)\frac{\text{dx}}{\text{dy}}=1$
$\Rightarrow\ \Big(\frac{\text{e}^{-2\sqrt{\text{x}}}}{\sqrt{\text{x}}}-\frac{\text{y}}{\sqrt{\text{x}}}\Big)\frac{\text{dx}}{\text{dy}}=1$ $\Rightarrow\ \ \frac{\text{dy}}{\text{dx}}+\frac{\text{y}}{\sqrt{\text{x}}}=\frac{\text{e}^{-2\sqrt{\text{x}}}}{\sqrt{\text{x}}}$
$\text{Comparing this equation with }\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q},$ $\text{P}=\frac{1}{\sqrt{\text{x}}}\ \text{and}\ \text{Q}=\frac{\text{e}^{-2\sqrt{\text{x}}}}{\sqrt{\text{x}}}$
$\int\text{P}\ \text{dx}=\int\frac{1}{\sqrt{\text{x}}}\text{dx}=\int\text{x}^{\frac{-1}{2}}\text{dx}=\frac{\text{x}^{\frac{1}{2}}}{\frac{1}{2}}=2\sqrt{\text{x}}$ $\text{I.F}=\text{e}^{\int\text{pdx}}=\text{e}^{2\sqrt{\text{x}}}$
$\text{The general solution is}\ \ \text{y}(\text{I.F})=\int\text{Q}(\text{I.F})\text{dx}+\text{c}$
$\Rightarrow\ \ \text{y}^{2\sqrt{\text{x}}}=\int\frac{\text{e}^{-2\sqrt{\text{x}}}}{\sqrt{\text{x}}}\text{e}^{2\sqrt{\text{x}}}\ \text{dx}+\text{c}=\int\frac{1}{\sqrt{\text{x}}}\text{dx}+\text{x}$
$\Rightarrow\ \ \text{y}^{2\sqrt{\text{x}}}=\int\text{x}^{\frac{-1}{2}}\ \text{dx}+\text{c}=\frac{\text{x}^{\frac{1}{2}}}{\frac{1}{2}}+\text{c}=2\sqrt{\text{x}}+\text{c}$ $\Rightarrow\ \ \text{y}=\text{e}^{-2\sqrt{\text{x}}}\ (2\sqrt{\text{x}}+\text{c})$
View full question & answer→Question 363 Marks
In each of the verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:
$\text{y} = \text{x} \ \text{sin} \ \text{x}\ : \ \text{xy}' = \text{y}+\text{x}\sqrt{\text{x}^2-\text{y}^2} $ $(\text{x} \neq 0 \ \text{and} \ \text{x} > \text{y} \ \text{or} \ \text{x} < – \text{y})$
AnswerGiven: $y = x \sin x .....(i)$ To prove: $y$ given by eq. $(i)$ is a solution of differential equation $\text{xy}'=\text{y}+\text{x} \sqrt{\text{x}^2-\text{y}^2}\ ....(\text{ii})$
Proof: From eq. $(i),$
$\frac{\text{dy}}{\text{dx}}(=\text{y}') = \text{x} \frac{\text{d}}{\text{dx}}\text{sin}\ \text{x}+ \text{sin}\ \text{x} \frac{\text{d}}{\text{dx}} \text{x} = \text{x}\ \text{cos}\ \text{x}\ + {\text{sin}} \ \text{x}\text{L.H.S.}$ of eq. $(ii)$
$= xy' = x (x \cos x + \sin x) = x^2 \cos x + x \sin x \text{R.H.S.}$ of eq. $(ii)$
$= \text{y} + \text{x}\sqrt{\text{x}^2-\text{y}^2} = \text{x}\ \text{sin} \ \text{x}+\text{x} \sqrt{\text{x}^2-\text{x}^2\ \text{sin}^2\text{x}}$ [From eq. (i)]$= \text{x}\ \text{sin}\ \text{x}+ \text{x} \sqrt{\text{x}^2(1-\text{sin}^2\text{x})} = \text{x}\ \text{sin}\ \text{x}+\text{x}\sqrt{\text{x}^2\text{cos}^2\text{x}}$
$= x \sin x + x.x \cos x = x \sin x + x^2 \cos x$
$= x^2 \cos x + x \sin x$
$\therefore \text{L.H.S. = R.H.S}$
Hence, $y$ given by eq. $(i)$ is a solution of $\text{xy}'=\text{y}+\text{x}\sqrt{\text{x}^2-\text{y}^2}.$
View full question & answer→Question 373 Marks
For the following differntial equations verify that the accompanying function is a solution:
| Differential equation |
Function |
| $\text{x}^3\frac{\text{d}{^2}\text{y}}{\text{dx}^2}=1$ |
$\text{y}=\text{ax}+\text{b}+\frac{1}{2\text{x}}$ |
AnswerWe have $\text{y}=\text{ax}+\text{b}+\frac{1}{2\text{x}}\ ...(1)$ Differentiating both sides of (1) with respect to x, we get $\frac{\text{dy}}{\text{dx}}=\text{a}-\frac{1}{2\text{x}^2}\ ...(2)$ Now, differentiating both sides of (2) with respect to x, we get $\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\Big(-\frac{1}{2}\Big)\times\Big(\frac{-2}{\text{x}^3}\Big)$ $\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{1}{\text{x}^3}$ $\Rightarrow\text{x}^3\frac{\text{d}^2\text{y}}{\text{dx}^2}=1$Hence, the given function is the solution to the given differential equation.
View full question & answer→Question 383 Marks
For each of the differential equations in find a particular solution satisfying the given condition:
$\text{cos}\bigg(\frac{\text{dy}}{\text{dx}}\bigg) = \text{a}(\text{a} \in \text{R}); \ \text{y}=1$ when $x = 0$
AnswerGiven: Differention equation $\text{cos}\bigg(\frac{\text{dy}}{\text{dx}}\bigg) = \text{a}(\text{a} \in \text{R}); \ \text{y}=1$ when $x = 0$
$\ \Rightarrow \ \frac{\text{dy}}{\text{dx}}=\text{cos}^{-1}\text{a}$
$\Rightarrow \ \text{dy}=\Big(\text{cos}^{-1}\text{a}\Big)\text{dx}$
Integrating both sides, $\int 1\ \text{dy}=\int\Big(\text{cos}^{-1}\text{a}\Big)\text{dx}$
$\Rightarrow \ \text{y}=\Big(\text{cos}^{-1}\text{a}\Big)\int1\ \text{dx}$
$\ \Rightarrow \ \text{y}=\Big(\text{cos}^{-1}\text{a}\Big)\ \text{x+c}\ .....\text{(i)}$
Now putting $y = 1$ when $x = 0$ in eq. $(i),$ we get $c = 1$
Putting $c = 1$ in eq. $(i), y = (\cos ^{-1} a) x + 1$
$\Rightarrow\frac{\text{y-1}}{\text{x}}=\cos^{-1} \text{a}$
$\Rightarrow\ \cos\frac{\text{y-1}}{\text{x}}=\text{a}$
View full question & answer→Question 393 Marks
In each of the verify that the given function (explicit or implicit) is a solution of the corresponding differential equation: $\text{y} – \text{cos}\ \text{y} = \text{x} \ :\ (\text{y} \ \text{sin} \ \text{y} + \text{cos} \ \text{y} + \text{x}) \text{y}' = \text{y}$
AnswerGiven: y-cos y = x ....(i) To prove: y given by eq. (i) is a solution of differential equation(y sin y + cos y + x) y' = y ....(ii)
Proof: Differentiating both sides of eq. (i) w.r.t x, we have$\text{y}'+(\text{sin}\ \text{y})\text{y}'=1 \ \Rightarrow\ \ \text{y}'(1+\text{sin}\ \text{y})=1$
$\Rightarrow\ \ \text{y}'=\frac{1}{1+\text{sin}\ \text{y}} \ ....(\text{iii})$ Putting the value of x from eq. (i) and value of y' from eq. (iii) in L.H.S. of eq. (ii),$\text{(y sin y + cos y + x)}{\text{y}}' $ $\ \Rightarrow \ \text{(y sin y + cos y + y} -\text{cos} \ \text{y}) \frac{1}{1+\text{sin}\ \text{y}}$
$\Rightarrow \ \text{(y sin y + y)}\frac{1}{1+ \text{sin y}} $ $\ \Rightarrow \ \text{y}(\text{sin y} + 1)\frac{1}{1+\text{sin y}}= \text{y = R.H.S. of (ii)}$ Hence, Function given by eq. (i) is a solution of (y sin y + cos y + x) y' = y.
View full question & answer→Question 403 Marks
verify that $\text{y}=\text{cx}+2\text{c}^2$ is a solution of the differential equation $2\Big(\frac{\text{d}\text{y}}{\text{dx}}\Big)^2-\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}=0$
AnswerWe have,
$\text{y}=\text{cx}+2\text{c}^2\ ...(1)$
Differentiating both sides of (1) with respect to x, we get
$\frac{\text{dy}}{\text{dx}}=\text{c}\ ...(2)$
Now,
$2\Big(\frac{\text{dy}}{\text{dx}}\Big)^2+\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}$
$=2\text{c}^2+\text{cx}-\text{cx}-2\text{c}^2=0$
$\Rightarrow\Big(\frac{\text{d}\text{y}}{\text{dx}}\Big)^2+\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}=0$
Hence, the given is the solution to the given differential equation.
View full question & answer→Question 413 Marks
Solve the following differential equations:
$\text{xy}\frac{\text{dy}}{\text{dx}}=1+\text{x + y + xy}$
Answer$\text{xy}\frac{\text{dy}}{\text{dx}}=1+\text{x + y + xy}$
$=(1+\text{x})+\text{y}(1+\text{x})$
$\text{xy}\frac{\text{dy}}{\text{dx}}=(1+\text{x})(1+\text{y})$
$\int\frac{\text{ydy}}{\text{y}+1}=\int\frac{1+\text{x}}{\text{x}}\text{dx}$
$\int\Big(1-\frac{1}{\text{y}+1}\Big)\text{dy}=\int\Big(\frac{1}{\text{x}}+1\Big)\text{dx}$
$\text{y}-\log|\text{y}+1|=\log|\text{x}|+\text{x}+\log|\text{c}|$
$\text{y}=\log|\text{cx(y+1})|+\text{x}$
View full question & answer→Question 423 Marks
In each of the verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:
$\text{y}=\sqrt{\text{a}^2-\text{x}^2}\text{x} \in(-\text{a, a}) :\ \text{x+y}\frac{\text{dy}}{\text{dx}}=\ 0 (\text{y}\neq0)$
AnswerGiven: $\text{y}=\sqrt{\text{a}^2-\text{x}^2},\ \text{x}\in(-\text{a, a}) \ ....(\text{i})$ To prove: y given by eq. (i) is a solution of differential equation $\text{x+y}\frac{\text{dy}}{\text{dx}} = 0 \ ....(\text{ii})$ Proof: From eq. (i), $\frac{\text{dy}}{\text{dx}}=\frac{1}{2}(\text{a}^2-\text{x}^2)^\frac{1}{2}\frac{\text{d}}{\text{dx}}(\text{a}^2-\text{x}^2)=\frac{1}{2\sqrt{\text{a}^2-\text{x}^2}}(-2\text{x})$$= \frac{-\text{x}}{\sqrt{\text{a}^2-\text{x}^2}}\ .....(\text{iii})$
Putting the values of y and $\frac{\text{dy}}{\text{dx}}$ from eq. (i) and (iii) in L.H.S. of eq. (ii), $\text{x}+\text{y}\frac{\text{dy}}{\text{dx}}=\text{x}+\sqrt{\text{a}^2-\text{x}^2}\bigg(\frac{-\text{x}}{\sqrt{\text{a}^2-\text{x}^2}}\bigg)$ $=\text{x}-\text{x}=0=\text{R.H.S. of eq.(ii)}$ Hence, Function given by eq. (i) is a solution of $\text{x+y}\frac{\text{dy}}{\text{dx}}=0.$
View full question & answer→Question 433 Marks
Form the differential equation having $\text{y} = (\sin^{-1}\text{x})^{2} + \text{A}\cos^{-1} + \text{ B},$ where A and B are arbitrary constants, as its general solution.
AnswerWe have, $\text{y}=(\sin^{-1}\text{x})^2+\text{A}\cos^{-1}\text{x}+\text{B}$
On differentiating w.r.t.x, we get
$\frac{\text{dy}}{\text{dx}}=\frac{2\sin^{-1}\text{x}}{\sqrt{1-\text{x}^2}}+\frac{(-\text{A})}{\sqrt{1-\text{x}^2}}$
$\Rightarrow\sqrt{1-\text{x}^2}\frac{\text{dy}}{\text{dx}}=2\sin^{-1}\text{x}-\text{A}$
Again, differentiating w.r.t.x, we get
$\sqrt{1-\text{x}^2}\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}+\frac{\text{dy}}{\text{dx}}.\frac{-2\text{x}}{2\sqrt{1+\text{x}^2}}=\frac{2}{{\sqrt{1-\text{x}}^2}}$
$\Rightarrow(1-\text{x}^2)\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}-\frac{\text{x}}{\sqrt{1-\text{x}^2}}.\sqrt{1-\text{x}^2}\frac{\text{dy}}{\text{dx}}=2$
$\Rightarrow(1-\text{x}^2)\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}-\text{x}\frac{\text{dy}}{\text{dx}}=2$
$\Rightarrow(1-\text{x}^2)\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}-\text{x}\frac{\text{dy}}{\text{dx}}-2=0$
which is the required differential equation.
View full question & answer→Question 443 Marks
Solve the differential equation $\text{y e}^{\frac{\text{x}}{\text{y}}}\text{dx}=\Big(\text{x}\ \text{e}^{\frac{\text{x}}{\text{y}}}+\text{y}^2\Big)\text{dy}(\text{y}\neq0).$
AnswerGiven: Differential equation $\text{y.e}^{\frac{\text{x}}{\text{y}}}\text{dx}=\Big(\text{x}.\text{e}^{\frac{\text{x}}{\text{y}}}+\text{y}^2\Big)\text{dy},\ \text{y}\neq0$
$\Rightarrow\ \ \frac{\text{dx}}{\text{dy}}=\frac{\text{x.e}^{\frac{\text{x}}{\text{y}}}+\text{y}^2}{\text{y.e}^{\frac{\text{x}}{\text{y}}}}=\frac{\text{x.e}^{\frac{\text{x}}{\text{y}}}}{\text{y.e}^\frac{\text{x}}{\text{y}}}+\frac{\text{y}^2}{\text{y.e}^{\frac{\text{x}}{\text{y}}}}$ $\Rightarrow\ \ \frac{\text{dx}}{\text{dy}}=\frac{\text{x}}{\text{y}}+\text{y.e}^{\frac{-\text{x}}{\text{y}}}\ \ ...\text{(i)}$
It is not a homogeneous differential equation because of presence of only y as a factor, yet it can be solved by putting $\frac{\text{x}}{\text{y}}=\text{v},\ \text{i.e}\ \text{x}=\text{vy}.\ \ \Rightarrow\ \ \frac{\text{dx}}{\text{dy}}=\text{v}+\text{y}\frac{\text{dv}}{\text{dy}}$
Putting these values in eq. (i), we get
$\text{v}+\text{y}\frac{\text{dv}}{\text{dy}}=\text{v}+\text{y e}^{-\text{v}}\ \ \Rightarrow\ \ \text{y}\frac{\text{dv}}{\text{dy}}=\text{y.e}^{-\text{x}}$ $\Rightarrow\ \ \text{y}\frac{\text{dv}}{\text{dy}}=\frac{\text{y}}{\text{e}^{\text{v}}}$
$\Rightarrow\ \ \text{e}^\text{v}=\text{y}+\text{c}\ \ \Rightarrow\ \ \text{e}^{\frac{\text{x}}{\text{y}}}=\text{y}+\text{c}$
View full question & answer→Question 453 Marks
Find the general solution of the differential equation $\frac{\text{dy}}{\text{dx}}+\sqrt{\frac{1-\text{y}^2}{1-\text{x}^2}}=0.$
AnswerThe given differential equation is:
$\frac{\text{dy}}{\text{dx}}+\sqrt{\frac{1-\text{y}^2}{1-\text{x}^2}}=0$
$\text{or}\ \ \frac{\text{dy}}{\text{dx}}=-\sqrt{\frac{1-\text{y}^2}{1-\text{x}^2}},\ \ \text{or}\ \ \frac{1}{\sqrt{1-\text{y}^2}}\text{dy}=-\frac{1}{\sqrt{1-\text{x}^2}}\text{dx}$
Integrating both sides, w.r.t.x,
$\int\frac{1}{\sqrt{1-\text{y}^2}}\text{dy}=-\int\frac{1}{\sqrt{1-\text{x}^2}}\text{dx}$
$\therefore\ \sin^{-1}\text{y}=-\sin^{-1}\text{x}+\sin^{-1}\text{c, where }\sin^{-1}\text{c}$ $\text{is an arbitrary constant}.$
$\therefore\ \sin^{-1}\text{x}+\sin^{-1}\text{y}=\sin^{-1}\text{c}$
$\Rightarrow\ \ \sin^{-1}\Big[\text{x}\sqrt{1-\text{y}^2}+\text{y}\sqrt{1-\text{x}^2}\Big]=\sin^{-1}\text{c}$
$\text{or}\ \ \text{x}\Big[\sqrt{1-\text{y}^2}+\text{y}\sqrt{1-\text{x}^2}\Big]=\text{c,}$ $\text{where c is an arbitrary constant}.$
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For each of the differential equations in find the general solution:
y log y dx – x dy = 0
AnswerGiven: Differential equation
$\text{y log y dx}- \text{x dy=0}\ \Rightarrow \ -\text{x dy =}-\text{y log y dx}$
$\Rightarrow \ \frac{\text{dy}}{\text{y log y}}= \frac{\text{dx}}{\text{x}} \ \ [\text{Separating variables}]$
Integrating both sides, $\int \frac{\text{dy}}{\text{y log y}}=\int\frac{\text{dx}}{\text{x}}$
Putting log y=t on L.H.S., we get $\frac{1}{\text{y}}=\frac{\text{dt}}{\text{dy}}\ \Rightarrow\ \frac{\text{dy}}{\text{y}}=\text{dt}$
Now eq. (i) becomes $\ \int \frac{\text{dt}}{\text{t}}=\int\frac{\text{dx}}{\text{x}}$
$\Rightarrow \ \ \text{log}|\text{t}|=\text{log}|\text{x}|+\text{log}|\text{c}|=\text{log}|\text{xc}|$
[If all the terms in the solution of a differential equation involve log, it is better to use log c or log |c| instead of c in the solution.]
$\Rightarrow \ |\text{t}|=|\text{xc}|\ \Rightarrow \ \text{t}=\pm\text{xc}$ $\Rightarrow \ \text{log} \ \text{y}=\pm \text{xc}=\text{ax}\ \text{where a =}\ \pm\text{c}$
$\Rightarrow \ \text{y = e}^{\text{ax}}$
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For each of the differential equations in find the general solution:
$\frac{\text{dy}}{\text{dx}}=\sqrt{4-\text{y}^2}(-2<\text{y}<2)$
AnswerGiven: Differential equation $\frac{\text{dy}}{\text{dx}}=\sqrt{4-\text{y}^2} \ \Rightarrow \ \ \text{dy}=\sqrt{4-\text{y}^2}\ \text{dx}$
$\Rightarrow \ \frac{\text{dy}}{\sqrt{4-\text{y}^2}}=\text{dx}$
Integrating both sides, $\int \frac{\text{dy}}{\sqrt{4-\text{y}^2}}\ \text{dy}=\int1\ \text{dx}$
$\Rightarrow \ \ \text{sin}^{-1}\frac{\text{y}}{2}=\text{x}+\text{c}\ \ \Bigg[\therefore\int\frac{1}{\sqrt{\text{a}^2-\text{x}^2}}\ \text{dx}=\text{sin}^{-1}\frac{\text{x}}{\text{a}}\Bigg]$
$\frac{\text{y}}{2}=\text{sin}(\text{x+c)} \ \Rightarrow \ \text{y}=2\text{sin}(\text{x+c)}$
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Find a particular solution of the differential equation $\frac{\text{dy}}{\text{dx}}+\text{y}\cot\text{x}=4\text{x}\ \text{cosec}\ \text{x}\ (\text{x}\neq0),$ $\text{given that y}=0\ \text{when x}=\frac{\pi}{2}.$
AnswerGiven: Differential equation $\frac{\text{dy}}{\text{dx}}+\text{y}\cot\text{x}=4\text{x}\ \text{cosec}\ \text{x}$
Comparing this equation with $\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q},\ \ \text{P}=\cot\text{x}\ \text{and}\ \text{Q}=4\text{x}\ \text{cosec}\ \text{x}$
$\int\text{P}\ \text{dx}=\int\cot\text{x}\ \text{dx}=\log\sin\text{x}$ $\text{I.F}=\text{e}^{\int\text{p dx}}=\text{e}^{\log\sin\text{x}}=\sin\text{x}$
The general solution is $\text{y}(\text{I.F})=\int\text{Q}(\text{I.F}.)\ \text{dx}+\text{c}$
$\Rightarrow\ \ \text{y}(\sin\text{x})=\int4\text{x}\ \text{cosec}\ \text{x}\ \sin\text{x}\ \text{dx}+\text{c}$ $\Rightarrow\ \ \text{y}(\sin\text{x})=4\int\text{x}.\frac{1}{\sin\text{x}}\sin\text{x}\ \text{dx}+\text{c}$
$\Rightarrow\ \ \text{y}(\sin\text{x})=4\int\text{x}\ \text{dx}+\text{c}=4\frac{\text{x}^2}{2}+\text{c}$ $\Rightarrow\ \ \text{y}\sin\text{x}=2\text{x}^2+\text{c}\ \ ...{(\text{i})}$
$\text{Now Putting y}=0,\text{x}=\frac{\pi}{2}\ \text{in eq. (i)},$ $0=2.\frac{\pi^{2}}{4}+\text{c}\ \ \Rightarrow\ \ \text{c}=\frac{-\pi^{2}}{2}$
$\text{Putting c}=\frac{-\pi^2}{2}\ \text{in eq. (i)},\ \ \text{y}\sin\text{x}=2\text{x}^2-\frac{\pi^2}{2}$
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Solve the following differential equations:
$\sqrt{1+\text{x}^2}\ \text{dy}+\sqrt{1+\text{y}^2}\ \text{dx}=0$
Answer$\sqrt{1+\text{x}^2}\ \text{dy}+\sqrt{1+\text{y}^2}\ \text{dx}=0$
$\sqrt{1+\text{x}^2}\ \text{dy}=-\sqrt{1+\text{y}^2}\ \text{dx}$
$\int\frac{\text{dy}}{\sqrt{1+\text{y}^2}}=-\int\frac{\text{dx}}{\sqrt{1+\text{x}^2}}$
$\log|\text{y}+\sqrt{1+\text{y}^2}|=-\log|\text{x}+\sqrt{1+\text{x}^2}|=\log|\text{c}|$
$(\text{y}+\sqrt{1+\text{y}^2})(\text{x}+\sqrt{1+\text{x}^2})=\text{c}$
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Solve the following equation
$\text{xy dy}=(\text{y}-1)(\text{x}+1)\text{dx}$
Answer$\text{xy dy}=(\text{y}-1)(\text{x}+1)\text{dx}$
$\frac{\text{y}}{\text{y}-1}\text{dy}=\frac{\text{x}+1}{\text{x}}\ \text{dx}$
$\int\Big(1+\frac{1}{\text{y}-1}\Big)\text{dy}=\int\Big(1+\frac{1}{\text{x}}\Big)\text{dx}$
$\text{y}+\log|\text{y}-1|=\text{x}+\log|\text{x}|+\text{C}$
$\text{y}-\text{x}=\log|\text{x}|-\log|\text{y}-1|+\text{C}$
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