NCERT Solutions for Miscellaneous Exercise Chapter 9 Class 12 - Differential Equations

Question 1: Indicate Order and Degree.

(i) $\frac{d^2y}{dx^2} + 5x \left (\frac{dy}{dx} \right )^2-6y = \log x$

Answer:

Given function is
$\frac{d^2y}{dx^2} + 5x \left (\frac{dy}{dx} \right )^2-6y = \log x$
We can rewrite it as
$y''+5x(y')^2-6y = \log x$
Now, it is clear from the above that, the highest order derivative present in differential equation is $y''$

Therefore, the order of the given differential equation $\frac{d^2y}{dx^2} + 5x \left (\frac{dy}{dx} \right )^2-6y = \log x$ is 2
Now, the given differential equation is a polynomial equation in its derivative y '' and y 'and power raised to y '' is 1
Therefore, it's degree is 1

Question 1: Indicate Order and Degree.

(ii) $\left(\frac{dy}{dx} \right )^3 - 4\left(\frac{dy}{dx} \right )^2 + 7y = \sin x$

Answer:

Given function is
$\left(\frac{dy}{dx} \right )^3 - 4\left(\frac{dy}{dx} \right )^2 + 7y = \sin x$
We can rewrite it as
$(y')^3-4(y')^2+7y=\sin x$
Now, it is clear from the above that, the highest order derivative present in differential equation is y'

Therefore, order of given differential equation is 1
Now, the given differential equation is a polynomial equation in it's dervatives y 'and power raised to y ' is 3
Therefore, it's degree is 3

Question 1: Indicate Order and Degree.

(iii) $\frac{d^4 y}{dx^4} - \sin\left(\frac{d^3y}{dx^3} \right ) = 0$

Answer:

Given function is
$\frac{d^4 y}{dx^4} - \sin\left(\frac{d^3y}{dx^3} \right ) = 0$
We can rewrite it as
$y''''-\sin y''' = 0$
Now, it is clear from the above that, the highest order derivative present in differential equation is y''''

Therefore, order of given differential equation is 4
Now, the given differential equation is not a polynomial equation in it's dervatives
Therefore, it's degree is not defined

Question 2: Verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.

(i) $xy = ae^x + be^{-x} + x^2\qquad :\ x\frac{d^2y}{dx^2} + 2\frac{dy}{dx} - xy +x^2 -2 =0$

Answer:

Given,

$xy = ae^x + be^{-x} + x^2$

Now, differentiating both sides w.r.t. x,

$x\frac{dy}{dx} + y = ae^x - be^{-x} + 2x$

Again, differentiating both sides w.r.t. x,

$\\ (x\frac{d^2y}{dx^2} + \frac{dy}{dx}) + \frac{dy}{dx} = ae^x + be^{-x} + 2 \\ $

$\implies x\frac{d^2y}{dx^2} + 2\frac{dy}{dx} = ae^x + be^{-x} + 2 \\ $

$\implies x\frac{d^2y}{dx^2} + 2\frac{dy}{dx} = xy -x^2 + 2 \\$

$\implies x\frac{d^2y}{dx^2} + 2\frac{dy}{dx} - xy + x^2 + 2$

Therefore, the given function is the solution of the corresponding differential equation.

Question 2: Verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.

(ii) $y = e^x(a\cos x + b \sin x )\qquad : \ \frac{d^2y}{dx^2} - 2\frac{dy}{dx} + 2y = 0$

Answer:

Given,

$y = e^x(a\cos x + b \sin x )$

Now, differentiating both sides w.r.t. x,

$\frac{dy}{dx} = e^x(-a\sin x + b \cos x ) + e^x(a\cos x + b \sin x ) =e^x(-a\sin x + b \cos x ) +y$

Again, differentiating both sides w.r.t. x,

$\\ \frac{d^2y}{dx^2} = e^x(-a\cos x - b \sin x ) + e^x(-a\sin x + b \cos x ) + \frac{dy}{dx} \\$

$= -y + (\frac{dy}{dx} -y) + \frac{dy}{dx} \\$

$\implies \frac{d^2y}{dx^2} - 2\frac{dy}{dx} + 2y = 0$

Therefore, the given function is the solution of the corresponding differential equation.

Question 2 Verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.

(iii) $y= x\sin 3x \qquad : \ \frac{d^2y}{dx^2} + 9y - 6\cos 3x = 0$

Answer:

Given,

$y= x\sin 3x$

Now, differentiating both sides w.r.t. x,

$y= x\sin 3x \frac{dy}{dx} = x(3\cos 3x) + \sin 3x$

Again, differentiating both sides w.r.t. x,

$\\ \frac{d^2y}{dx^2} = 3x(-3\sin 3x) + 3\cos 3x + 3\cos 3x \\ $

$= -9y + 6\cos 3x \\$

$\implies \frac{d^2y}{dx^2} + 9y - 6\cos 3x = 0$

Therefore, the given function is the solution of the corresponding differential equation.

Question 2: Verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.

(iv) $x^2 = 2y^2\log y\qquad : \ (x^2 + y^2)\frac{dy}{dx} - xy = 0$

Answer:

Given,

$x^2 = 2y^2\log y$

Now, differentiating both sides w.r.t. x,

$\\ 2x = (2y^2.\frac{1}{y} + 2(2y)\log y)\frac{dy}{dx} = 2(y + 2y\log y)\frac{dy}{dx} \\$

$\implies \frac{dy}{dx} = \frac{x}{y(1+ 2\log y)}$

Putting $\frac{dy}{dx}\ and \ x^2$ values in LHS

$\\ (2y^2\log y + y^2)\frac{dy}{dx} - xy = y^2(2\log y + 1)\frac{x}{y(1+ 2\log y)} -xy \\$

$= xy - xy = 0 = RHS$

Therefore, the given function is the solution of the corresponding differential equation.

Question 3: Form the differential equation representing the family of curves given by $(x-a)^2 + 2y^2 = a^2$ , where a is an arbitrary constant.

Answer:

Given equation is
$(x-a)^2 + 2y^2 = a^2$
we can rewrite it as
$2y^2 = a^2-(x-a)^2$ -(i)
Differentiate both the sides w.r.t x
$\frac{d\left ( 2y^2 \right )}{dx}=\frac{d(a^2-(x-a)^2)}{dx}$
$4yy^{'}=4y\frac{dy}{dx}=-2(x-a)\\ \\$
$(x-a)= -2yy'\Rightarrow a = x+2yy'$ -(ii)
Put value from equation (ii) in (i)
$(-2yy')^2+2y^2= (x+2yy')^2\\$

$4y^2(y')^2+2y^2= x^2+4y^2(y')^2+4xyy'\\$

y' = \frac{2y^2-x^2}{4xy}$
Therefore, the required differential equation is $y' = \frac{2y^2-x^2}{4xy}$

Question 4: Prove that $x^2 - y^2 = c (x^2 + y^2 )^2$ is the general solution of differential equation $(x^3 - 3x y^2 ) dx = (y^3 - 3x^2 y) dy$ , where c is a parameter.

Answer:

Given,

$
\begin{aligned}
& \left(x^3-3 x y^2\right) d x=\left(y^3-3 x^2 y\right) d y \\
& \Longrightarrow \frac{d y}{d x}=\frac{\left(x^3-3 x y^2\right)}{\left(y^3-3 x^2 y\right)}
\end{aligned}
$

Now, let $\mathbf{y}=\mathrm{vx}$

$
\Longrightarrow \frac{d y}{d x}=\frac{d(v x)}{d x}=v+x \frac{d v}{d x}
$

Substituting the values of $y$ and $y^{\prime}$ in the equation,

$
\begin{aligned}
& v+x \frac{d v}{d x}=\frac{\left(x^3-3 x(v x)^2\right)}{\left((v x)^3-3 x^2(v x)\right)} \\
& \Longrightarrow v+x \frac{d v}{d x}=\frac{1-3 v^2}{v^3-3 v} \\
& \Longrightarrow x \frac{d v}{d x}=\frac{1-3 v^2}{v^3-3 v}-v=\frac{1-v^4}{v^3-3 v} \\
& \Longrightarrow\left(\frac{v^3-3 v}{1-v^4}\right) d v=\frac{d x}{x}
\end{aligned}
$

Integrating both sides we get,

$
\int\left(\frac{v^3-3 v}{1-3 v^4}\right) d v=\log x+\log C^{\prime}
$

Now, $\int\left(\frac{v^3-3 v}{1-3 v^4}\right) d v=\int \frac{v^3}{1-v^4} d v-3 \int \frac{v d v}{1-v^4}$

$
\Rightarrow \int\left(\frac{v^3-3 v}{1-3 v^4}\right) d v=I_1-3 I_2 \text {, where } I_1=\int \frac{v^3}{1-v^4} d v \text { and } I_2=\int \frac{v d v}{1-v^4}
$

Let $1-v^4=\mathrm{t}$

$
\begin{aligned}
& \frac{d}{d v}\left(1-v^4\right)=\frac{d t}{d v} \\
& \Longrightarrow-4 v^3=\frac{d t}{d v} \\
& \Longrightarrow v^3 d v=-\frac{d t}{4}
\end{aligned}
$

Now,

$
\mathrm{I}_1=\int-\frac{\mathrm{dt}}{4}=-\frac{1}{4} \log \mathrm{t}=-\frac{1}{4} \log \left(1-\mathrm{v}^4\right)
$

and

$
I_2=\int \frac{v d v}{1-v^4}=\int \frac{v d v}{1-\left(v^2\right)^2}
$

Let $v^2=p$

$
\begin{aligned}
& \Rightarrow \frac{d}{d v}\left(v^2\right)=\frac{d p}{d v} \\
& \Rightarrow 2 v=\frac{d p}{d v}
\end{aligned}
$

$
\Rightarrow \mathrm{vdv}=\frac{\mathrm{dp}}{2}
$

$
\therefore \mathrm{I}_2=\frac{1}{2} \int \frac{\mathrm{dp}}{1-\mathrm{p}^2}=\frac{1}{2 \times 2} \log \left|\frac{1+\mathrm{p}}{1-\mathrm{p}}\right|=\frac{1}{4}\left|\frac{1+\mathrm{v}^2}{1-\mathrm{v}}\right|
$

Now, substituting the values of $\mathrm{I}_1$ and $\mathrm{I}_2$ in the above equation, we get,

$
\int\left(\frac{\mathrm{v}^3-3 \mathrm{y}}{1-\mathrm{v}^4}\right) \mathrm{dv}=-\frac{1}{4} \log \left(1-\mathrm{v}^4\right)-\frac{3}{4} \log \left|\frac{1+\mathrm{v}^2}{1-\mathrm{v}^2}\right|
$

Thus,

$
\begin{aligned}
& -\frac{1}{4} \log \left(1-\mathrm{v}^4\right)-\frac{3}{4} \log \left|\frac{1+\mathrm{v}^2}{1-\mathrm{v}^2}\right|=\log \mathrm{x}+\log \mathrm{C}^{\prime} \\
& \Rightarrow-\frac{1}{4} \log \left[\left(1-\mathrm{v}^4\right)\left(\frac{1+\mathrm{v}^2}{1-\mathrm{v}^2}\right)^3\right]=\log \mathrm{C}^{\prime} \mathrm{x} \\
& \Rightarrow \frac{\left(1+\mathrm{v}^2\right)^4}{\left(1-\mathrm{v}^2\right)^2}=\left(\mathrm{C}^{\prime} \mathrm{x}\right)^{-4} \\
& \Rightarrow \frac{\left(1+\frac{y^2}{x^2}\right)^4}{\left(1-\frac{y^2}{x^2}\right)^2}=\frac{1}{\mathrm{C}^4 \mathrm{x}^4} \\
& \left(x^2-y^2\right)^2=C^{\prime 4}\left(x^2+y^2\right)^4 \\
& \Longrightarrow\left(x^2-y^2\right)=C^{\prime 2}\left(x^2+y^2\right)^2 \\
& \Longrightarrow\left(x^2-y^2\right)=K\left(x^2+y^2\right)^2, \text { where } K=C^{\prime 2}
\end{aligned}
$

Question 5: Form the differential equation of the family of circles in the first quadrant which touch the coordinate axes.

Answer:

Now, equation of the circle with center at (x,y) and radius r is
$(x-a)^2+(y-b)^2 = r^2$
Since, it touch the coordinate axes in first quadrant
Therefore, x = y = r
$(x-a)^2+(y-a)^2 = a^2$ -(i)
Differentiate it w.r.t x
we will get
$2(x-a)+2(y-a)y'= 0\\ $

$\\ 2x-2a+2yy'-2ay' = 0\\$

$a=\frac{x+yy'}{1+y'}$ -(ii)
Put value from equation (ii) in equation (i)
$(x-\frac{x+yy'}{1+y'})^2+(y-\frac{x+yy'}{1+y'})^2=\left ( \frac{x+yy'}{1+y'} \right )^2\\$

$\\ (x+xy'-x-yy')^2+(y+yy'-x-yy')^2=(x+yy')^2\\ $

$\\ (y')^2(x-y)^2+(x-y)^2=(x+yy')^2\\$

$\\ (x-y)^2\left ( (y')^2+1 \right )=(x+yy')^2$
Therefore, the differential equation of the family of circles in the first quadrant which touches the coordinate axes is $(x-y)^2\left ( (y')^2+1 \right )=(x+yy')^2$

Question 6: Find the general solution of the differential equation $\frac{dy}{dx} + \sqrt{\frac{1 - y^2}{1-x^2}} = 0$

Answer:

Given equation is
$\frac{dy}{dx} + \sqrt{\frac{1 - y^2}{1-x^2}} = 0$
we can rewrite it as
$\frac{dy}{dx } =- \sqrt{\frac{1-y^2}{1-x^2}}\\$

$\\ \frac{dy}{\sqrt{1-y^2}}= \frac{-dx}{\sqrt{1-x^2}}$
Now, integrate on both the sides
$\sin^{-1}y + C =- \sin ^{-1}x + C'\\ $

$\\ \sin^{-1}y+\sin^{-1}x= C$
Therefore, the general solution of the differential equation $\frac{dy}{dx} + \sqrt{\frac{1 - y^2}{1-x^2}} = 0$ is $\sin^{-1}y+\sin^{-1}x= C$

Question 7: Show that the general solution of the differential equation $\frac{dy}{dx} + \frac{y^2 + y + 1}{x^2 + x + 1} = 0$ is given by $(x + y + 1) = A (1 - x - y - 2xy)$ , where A is parameter.

Answer:

Given,

$
\begin{aligned}
& \frac{d y}{d x}+\frac{y^2+y+1}{x^2+x+1}=0 \\
& \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=-\left(\frac{\mathrm{y}^2+\mathrm{y}+1}{\mathrm{x}^2+\mathrm{x}+1}\right) \\
& \Rightarrow \frac{\mathrm{dy}}{\mathrm{y}^2+\mathrm{y}+1}=\frac{-\mathrm{dx}}{\mathrm{x}^2 \mathrm{dx}+1} \\
& \Rightarrow \frac{\mathrm{dy}}{\mathrm{y}^2+\mathrm{y}+1}+\frac{\mathrm{dx}}{\mathrm{x}^2+\mathrm{x}+1}=0
\end{aligned}
$

Integrating both sides,

$
\begin{aligned}
& \int \frac{d y}{y^2+y+1}+\int \frac{d x}{x^2+x+1}=C \\
& \Rightarrow \int \frac{d y}{\left(y+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}+\int \frac{d y}{\left(x+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}=C \\
& \Rightarrow \frac{2}{\sqrt{3}} \tan ^{-1}\left[\frac{\mathrm{y}+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right]+\frac{2}{\sqrt{3}} \tan ^{-1}\left[\frac{x+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right]=C \\
& \Rightarrow \tan ^{-1}\left[\frac{2 y+1}{\sqrt{3}}\right]+\tan ^{-1}\left[\frac{2 x+1}{\sqrt{3}}\right]=\mathrm{C} \Rightarrow \tan ^{-1}\left[\frac{\frac{2 y+1}{\sqrt{3}}+\frac{2 x+1}{\sqrt{3}}}{1-\frac{2 y+1}{\sqrt{3}} \cdot \frac{2 x+1}{\sqrt{3}}}\right]=\frac{\sqrt{3}}{2} C \\
& \Rightarrow \tan ^{-1}\left[\frac{\frac{2 x+2 y+2}{\sqrt[3]{3}}}{1-\left(\frac{4 x y+2 x+2 y+1}{3}\right)}\right]=\frac{\sqrt{3}}{2} C \\
& \Rightarrow \tan ^{-1}\left[\frac{2 \sqrt{3}(x+y+1)}{3-4 x y-2 x-2 y-1}\right]=\frac{\sqrt{3}}{2} C \\
& \Rightarrow \tan ^{-1}\left[\frac{2 \sqrt{3}(x+y+1)}{2(1-x-y-2 x y)}\right]=\frac{\sqrt{3}}{2} C \\
& \Rightarrow \frac{\sqrt{3}(x+y+1)}{(1-x-y-2 x y)}=\tan \left(\frac{\sqrt{3}}{2} c\right)
\end{aligned}
$

Let $\tan \left(\frac{\sqrt{3}}{2} c\right)=B$

$
x+y+1=\frac{2 B}{\sqrt{3}}(1-x-y-2 x y)
$

Let $A=\frac{2 B}{\sqrt{3}}$,

$
x+y+1=A(1-x-y-2 x y)
$

Hence proved.

Question 8: Find the equation of the curve passing through the point $\left(0,\frac{\pi}{4} \right )$ whose differential equation is $\sin x \cos y dx + \cos x \sin y dy = 0.$

Answer:

Given equation is
$\sin x \cos y dx + \cos x \sin y dy = 0.$
we can rewrite it as
$\frac{dy}{dx}= -\tan x\cot y\\$

$\\ \frac{dy}{\cot y}= -\tan xdx\\$

$\\ \tan y dy =- \tan x dx$
Integrate both the sides
$\log |\sec y|+C' = -\log|sec x|- C''\\ $

$\\ \log|\sec y | +\log|\sec x| = C\\$

$\\ \sec y .\sec x = e^{C}$
Now by using boundary conditiond, we will find the value of C
It is given that the curve passing through the point $\left(0,\frac{\pi}{4} \right )$
So,
$\sec \frac{\pi}{4} .\sec 0 = e^{C}\\ $

$\\ \sqrt2.1= e^C\\$

$\\ C = \log \sqrt2$
Now,
$\sec y.\sec x= e^{\log \sqrt 2}\\$

$\\ \frac{\sec x}{\cos y} = \sqrt 2\\ $

$\\ \cos y = \frac{\sec x}{\sqrt 2}$
Therefore, the equation of the curve passing through the point $\left(0,\frac{\pi}{4} \right )$ whose differential equation is $\sin x \cos y dx + \cos x \sin y dy = 0.$ is $\cos y = \frac{\sec x}{\sqrt 2}$

Question 9: Find the particular solution of the differential equation $(1 + e^ {2x} ) dy + (1 + y^2 ) e^x dx = 0$ , given that $y = 1$ when $x = 0$ .

Answer:

Given equation is
$(1 + e^ {2x} ) dy + (1 + y^2 ) e^x dx = 0$
we can rewrite it as
$\frac{dy}{dx}= -\frac{(1+y^2)e^x}{(1+e^{2x})}\\ $

$\\ \frac{dy}{1+y^2}= \frac{-e^xdx}{1+e^{2x}}$
Now, integrate both the sides
$\tan^{-1}y + C' =\int \frac{-e^{x}dx}{1+e^{2x}}$
$\int \frac{-e^{x}dx}{1+e^{2x}}\\$
Put
$e^x = t \\$

$e^xdx = dt$
$\int \frac{dt}{1+t^2}= \tan^{-1}t + C''$
Put $t = e^x$ again
$\int \frac{-e^{x}dx}{1+e^{2x}} = -\tan ^{-1}e^x+C''$
Put this in our equation
$\tan^{-1}y = -\tan ^{-1}e^x+C\\$

$\tan^{-1}y +\tan ^{-1}e^x=C$
Now, by using boundary conditions we will find the value of C
It is given that
y = 1 when x = 0
$\\ \tan^{-1}1 +\tan ^{-1}e^0=C\\ $

$\\ \frac{\pi}{4}+\frac{\pi}{4}= C\\$

$C = \frac{\pi}{2}$
Now, put the value of C

$\tan^{-1}y +\tan ^{-1}e^x=\frac{\pi}{2}$
Therefore, the particular solution of the differential equation $(1 + e^ {2x} ) dy + (1 + y^2 ) e^x dx = 0$ is $\tan^{-1}y +\tan ^{-1}e^x=\frac{\pi}{2}$

Question 10: Solve the differential equation $ye^\frac{x}{y}dx = \left(xe^\frac{x}{y} + y^2 \right )dy\ (y \neq 0)$

Answer:

Given,

$ye^\frac{x}{y}dx = (xe^\frac{x}{y} + y^2)dy$

$\\ ye^\frac{x}{y}\frac{dx}{dy} = xe^\frac{x}{y} + y^2 \\$

$\implies e^\frac{x}{y}[y\frac{dx}{dy} -x] = y^2 \\$

$\implies \frac{e^\frac{x}{y}[y\frac{dx}{dy} -x]}{y^2} = 1$

Let $\large e^\frac{x}{y} = t$

Differentiating it w.r.t. y, we get,

$\\ \frac{d}{dy}e^\frac{x}{y} = \frac{dt}{dy} \\$
$ \implies e^\frac{x}{y}.\frac{d}{dy}(\frac{x}{y}) = \frac{dt}{dy} \\$

$\implies \frac{e^\frac{x}{y}[y\frac{dx}{dy} -x]}{y^2} =\frac{dt}{dy}$

Thus from these two equations,we get,

$\\ \frac{dt}{dy} = 1 \\ $

$\implies \int dt = \int dy \\ $

$\implies t = y + C$

$\Longrightarrow e^{\frac{x}{y}}=y+C$

Question 11: Find a particular solution of the differential equation $(x - y) (dx + dy) = dx - dy,$ , given that $y = -1$ , when $x = 0$ . (Hint: put $x - y = t$ )

Answer:

Given equation is
$(x - y) (dx + dy) = dx - dy,$
Now, integrate both the sides
Put
$(x-y ) = t\\ $

$dx - dy = dt$
Now, given equation become
$dx+dy= \frac{dt}{t}$
Now, integrate both the sides
$x+ y + C '= \log t + C''$
Put $t = x- y$ again
$x+y = \log (x-y)+ C$
Now, by using boundary conditions we will find the value of C
It is given that
y = -1 when x = 0
$0+(-1) = \log (0-(-1))+ C\\$

$C = -1$
Now, put the value of C

$x+y = \log |x-y|-1\\ $

$\log|x-y|= x+y+1$
Therefore, the particular solution of the differential equation $(x - y) (dx + dy) = dx - dy,$ is $\log|x-y|= x+y+1$

Question 12 Solve the differential equation $\left[\frac{e^{-2\sqrt x}}{\sqrt x} - \frac{y}{\sqrt x} \right ]\frac{dx}{dy} = 1\; \ (x\neq 0)$ .

Answer:

Given,

$
\begin{aligned}
& {\left[\frac{e^{-2 \sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}}\right] \frac{d x}{d y}=1} \\
& \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{e}^{-2 \sqrt{x}}}{\sqrt{\mathrm{x}}}-\frac{\mathrm{y}}{\sqrt{x}} \\
& \Longrightarrow \frac{\mathrm{dy}}{\mathrm{dx}}+\frac{\mathrm{y}}{\sqrt{\mathrm{x}}}=\frac{\mathrm{e}^{-2 \sqrt{x}}}{\sqrt{\mathrm{x}}}
\end{aligned}
$

This is equation is in the form of $\frac{d y}{d x}+p y=Q$

$
p=\frac{1}{\sqrt{x}} \text { and } Q=\frac{e^{-2 \sqrt{x}}}{\sqrt{x}}
$

Now, I.F. $=\mathrm{e}^{\int \mathrm{pdx}}=\mathrm{e}^{\int \frac{1}{\sqrt{x}} \mathrm{dx}}=\mathrm{e}^{2 \sqrt{\mathrm{x}}}$
We know that the solution of the given differential equation is:

$
\begin{aligned}
& y(I \cdot F \cdot)=\int(Q \cdot F \cdot) d x+C \\
& \Rightarrow \mathrm{ye}^{2 \sqrt{\mathrm{x}}}=\int\left(\frac{\mathrm{e}^{-2 \sqrt{x}}}{\sqrt{\mathrm{x}}} \times \mathrm{e}^{2 \sqrt{\mathrm{x}}}\right) \mathrm{dx}+C \\
& \Rightarrow \mathrm{ye}^{2 \sqrt{\mathrm{x}}}=\int \frac{1}{\sqrt{\mathrm{x}}} \mathrm{dx}+\mathrm{C} \\
& \Rightarrow \mathrm{ye}^{2 \sqrt{x}}=2 \sqrt{\mathrm{x}}+C
\end{aligned}
$

Question 13: Find a particular solution of the differential equation $\frac{dy}{dx} + y \cot x = 4x \textup{cosec} x\ (x\neq 0)$ , given that $y = 0 \ \textup{when}\ x = \frac{\pi}{2}$ .

Answer:

Given equation is
$\frac{dy}{dx} + y \cot x = 4x \textup{cosec} x\ (x\neq 0)$
This is $\frac{dy}{dx} + py = Q$ type where $p =\cot x$ and $Q = 4xcosec x$ $Q = 4x \ cosec x$
Now,
$I.F. = e^{\int pdx}= e^{\int \cot xdx}= e^{\log |\sin x|}= \sin x$
Now, the solution of given differential equation is given by relation
$y(I.F.) =\int (Q\times I.F.)dx +C$
$y(\sin x ) =\int (\sin x\times 4x \ cosec x)dx +C$
$y(\sin x) =\int(\sin x\times \frac{4x}{\sin x}) +C\\$

$\\ y(\sin x) = \int 4x + C\\ $

$y\sin x= 2x^2+C$
Now, by using boundary conditions we will find the value of C
It is given that y = 0 when $x= \frac{\pi}{2}$
at $x= \frac{\pi}{2}$
$0.\sin \frac{\pi}{2 } = 2.\left ( \frac{\pi}{2} \right )^2+C\\$

$\\ C = - \frac{\pi^2}{2}$
Now, put the value of C
$y\sin x= 2x^2-\frac{\pi^2}{2}$
Therefore, the particular solution is $y\sin x= 2x^2-\frac{\pi^2}{2}, (sinx\neq0)$

Question 14: Find a particular solution of the differential equation $(x+1)\frac{dy}{dx} = 2e^{-y} -1$ , given that $y = 0$ when $x = 0$

Answer:

Given equation is
$(x+1)\frac{dy}{dx} = 2e^{-y} -1$
we can rewrite it as
$\frac{e^ydy}{2-e^y}= \frac{dx}{x+1}\\$
Integrate both the sides
$\int \frac{e^ydy}{2-e^y}= \log |x+1|\\$
$\int \frac{e^ydy}{2-e^y}$
Put
$2-e^y = t\\$

$-e^y dy = dt$
$\int \frac{-dt}{t}=- \log |t|$
put $t = 2- e^y$ again
$\int \frac{e^ydy}{2-e^y} =- \log |2-e^y|$
Put this in our equation
$\log |2-e^y| + C'= \log|1+x| + C''\\$

$\log (2-e^y)^{-1}= \log (1+x)+\log C\\$

$\frac{1}{2-e^y}= C(1+x)$

Now, by using boundary conditions we will find the value of C
It is given that y = 0 when x = 0
at x = 0
$\frac{1}{2-e^0}= C(1+0)\\ C = \frac{1}{2}$
Now, put the value of C
$\frac{1}{2-e^y} = \frac{1}{2}(1+x)\\ $'

$\\ \frac{2}{1+x}= 2-e^y\\$

$\frac{2}{1+x}-2= -e^y\\$

$-\frac{2x-1}{1+x} = -e^y\\$

$y = \log \frac{2x-1}{1+x}$
Therefore, the particular solution is $y = \log \frac{2x-1}{1+x}, x\neq-1$

Question 15: The population of a village increases continuously at the rate proportional to the number of its inhabitants present at any time. If the population of the village was 20, 000 in 1999 and 25000 in the year 2004, what will be the population of the village in 2009?

Answer:

Let n be the population of the village at any time t.

According to question,

$\frac{dn}{dt} = kn\ \ (k\ is\ a\ constant)$

$\\ \implies \int \frac{dn}{n} = \int kdt \\$

$\implies \log n = kt + C$

Now, at t=0, n = 20000 (Year 1999)

$\\ \implies \log (20000) = k(0) + C \\$

$\implies C = \log2 + 4$

Again, at t=5, n= 25000 (Year 2004)

$\\ \implies \log (25000) = k(5) + \log2 + 4 \\$

$\implies \log 25 + 3 = 5k + \log2 +4 \\$

$\implies 5k = \log 25 - \log2 -1 =\log \frac{25}{20} \\$

$\implies k = \frac{1}{5}\log \frac{5}{4}$

Using these values, at t =10 (Year 2009)

$\\ \implies \log n = k(10)+ C \\$

$\implies \log n = \frac{1}{5}\log \frac{5}{4}(10) + \log2 + 4 \\$

$\implies \log n = \log(\frac{25.2.10000}{16}) = \log(31250) \\$

$\therefore n = 31250$

Therefore, the population of the village in 2009 will be 31250.

Question 16: The general solution of the differential equation $\frac{ydx - xdy}{y} = 0$ is

(A) $xy = C$

(B) $x = Cy^2$

(C) $y = Cx$

(D) $y = Cx^2$

Answer:

Given equation is
$\frac{ydx - xdy}{y} = 0$
we can rewrite it as
$dx = \frac{x}{y}dy\\$

$\frac{dy}{y}=\frac{dx}{x}$
Integrate both the sides
we will get
$\log |y| = \log |x| + C\\ $

\log \frac{y}{x} = C \\ $

$\frac{y}{x} = e^C\\$

$\frac{y}{x} = C\\$

$y = Cx$
Therefore, answer is (C)

Question 17: The general solution of a differential equation of the type $\frac{dx}{dy} + P_1 x = Q_1$ is

(A) $ye^{\int P_1 dy} = \int \left(Q_1 e^{\int P_1 dy} \right )dy +C$

(B) $ye^{\int P_1 dx} = \int \left(Q_1 e^{\int P_1 dx} \right )dx +C$

(C) $xe^{\int P_1 dy} = \int \left(Q_1 e^{\int P_1 dy} \right )dy +C$

(D) $xe^{\int P_1 dx} = \int \left(Q_1 e^{\int P_1 dx} \right )dx +C$

Answer:

Given equation is
$\frac{dx}{dy} + P_1 x = Q_1$
and we know that the general equation of such type of differential equation is

$xe^{\int p_1dy} = \int (Q_1e^{\int p_1dy})dy+ C$
Therefore, the correct answer is (C)

Question 18: The general solution of the differential equation $e^x dy + (y e^x + 2x) dx = 0$ is

(A) $xe^y + x^2 = C$

(B) $xe^y + y^2 = C$

(C) $ye^x + x^2 = C$

(D) $ye^y + x^2 = C$

Answer:

Given equation is
$e^x dy + (y e^x + 2x) dx = 0$
we can rewrite it as
$\frac{dy}{dx}+y=-2xe^{-x}$
It is $\frac{dy}{dx}+py=Q$ type of equation where $p = 1 \ and \ Q = -2xe^{-x}$
Now,
$I.F. = e^{\int p dx }= e^{\int 1dx}= e^x$
Now, the general solution is
$y(I.F.) = \int (Q\times I.F.)dx+C$
$y(e^x) = \int (-2xe^{-x}\times e^x)dx+C\\$

$ye^x= \int -2xdx + C\\$

$ye^x=- x^2 + C\\ $

$ye^x+x^2 = C$
Therefore, (C) is the correct answer