Solution

The given differential equation is

Separating the variables, we get

$ \frac{d y}{d x}=2^{y-x} \Rightarrow \frac{d y}{d x}=\frac{2^{y}}{2^{x}} $

$ \frac{d y}{2^{y}}=\frac{d x}{2^{x}} \Rightarrow 2^{-y} d y=2^{-x} d x $

Integrating both sides, we get

$ \begin{aligned} \int 2^{-y} d y & =\int 2^{-x} d x \\ & \\ \frac{-2^{-y}}{\log 2} & =\frac{-2^{-x}}{\log 2}+c \quad \Rightarrow-2^{-y}=-2^{-x}+c \log 2 \\ \Rightarrow \quad-2^{-y}+2^{-x} & =c \log 2 \\ \Rightarrow \quad 2^{-x}-2^{-y} & =k \quad \quad \quad \text where \quad c \log 2=k \end{aligned} $

Solution

Equation of all non vertical lines are $y=m x+c$

Differentiating with respect to $x$, we get $\frac{d y}{d x}=m$

Again differentiating w.r.t. $x$ we have $\frac{d^{2} y}{d x^{2}}=0$

Hence, the required equation is $\frac{d^{2} y}{d x^{2}}=0$.

Solution

Given equation is

$ \begin{aligned} & \frac{d y}{d x} & =e^{-2 y} \end{aligned} $

Separating the variables, we get

$ \begin{aligned} \Rightarrow \quad \frac{d y}{e^{-2 y}} & =d x \Rightarrow e^{2 y} \cdot d y=d x \end{aligned} $

Integrating both sides, we get

$ \int e^{2 y} d y=\int d x \Rightarrow \frac{1}{2} e^{2 y}=x+c $

Put $y=0$ and $x=5$ (Given)

$\Rightarrow \frac{1}{2} e^{0}=5+c \Rightarrow c=\frac{1}{2}-5=-\frac{9}{2}$

$\therefore$ The equation becomes $\frac{1}{2} e^{2 y}=x-\frac{9}{2}$

Now putting $y=3$, we get

$ \frac{1}{2} e^{6}=x-\frac{9}{2} \Rightarrow x=\frac{1}{2} e^{6}+\frac{9}{2} $

Hence the required value of $x=\frac{1}{2}(e^{6}+9)$.

Solution

Given differential equation is

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

Dividing by $(x^{2}-1)$, we get

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

It is a linear differential equation of first order and first degree.

Here, $ P=\frac{2 x}{x^{2}-1}$ and $Q=\frac{1}{(x^{2}-1)^{2}}$

Integrating factor I.F. $=e^{\int P d x}=e^{\int \frac{2 x}{x^{2}-1} d x}=e^{\log |(x^{2}-1)|}=(x^{2}-1) , ((x^{2}-1)>0)$

$\therefore$ Solution of the equation is

$ \begin{aligned} & y \times \text{ I.F. }=\int Q \times \text{ I.F. } d x+\text{ C } \\ & \Rightarrow y \times(x^{2}-1)=\int \frac{1}{(x^{2}-1)^{2}} \times(x^{2}-1) d x+\text{ C } \\ & \Rightarrow y(x^{2}-1)=\int \frac{1}{x^{2}-1} d x+C \end{aligned} $

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

Hence the required solution is $y(x^{2}-1)=\frac{1}{2} \log (|\frac{x-1}{x+1}|)+C$.

Solution

Given equation is $\frac{d y}{d x}+2 x y=y$.

$ \Rightarrow \quad \frac{d y}{d x}=y-2 x y \quad \Rightarrow \frac{d y}{d x}=y(1-2 x) \\ \Rightarrow \frac{d y}{y}=(1-2 x) d x $

Integrating both sides, we have

$ \begin{aligned} & \int \frac{d y}{ y}=\int(1-2 x) d x \Rightarrow \log |y|=x-2 \cdot \frac{x^{2}}{2}+\log |c| \\ & \Rightarrow \quad \log |y|=x-x^{2}+\log |c| \Rightarrow \log |y|-\log |c|=x-x^{2} \\ & \Rightarrow \quad \log |\frac{y}{c}|=x-x^{2} \Rightarrow |{y}|=|{c}|e^{x-x^{2}} \\ \Rightarrow {y}=\pm {c}e^{x-x^{2}} \\ & \therefore \quad y=C \cdot e^{x-x^{2}} (where \quad C=\pm c) \end{aligned} $

Hence, the required solution is $y=C \cdot e^{x-x^{2}}$.

Solution

Given equation is $\frac{d y}{d x}+a y=e^{m x}$.

Here, $P=a$ and $Q=e^{m x}$

$\therefore$ I.F. $=e^{\int P d x}=e^{\int a \cdot d x}=e^{a x}$.

Solution of equation is $y \times$ I.F $=\int Q \times $ I.F. $d x+c$

$ \begin{matrix} \Rightarrow \quad y \cdot e^{a x}=\int e^{m x} \cdot e^{a x} d x+c \Rightarrow y \cdot e^{a x}=\int e^{(m+a) x} d x+c \\ \Rightarrow \qquad y \cdot e^{a x}=\frac{e^{(m+a) x}}{(m+a)}+c \Rightarrow y=\frac{e^{(m+a) x}}{(m+a)} \cdot e^{-a x}+c \cdot e^{-a x} \\ \therefore \qquad y=\frac{e^{m x}}{(m+a)}+c \cdot e^{-a x} \end{matrix} $

$ \Rightarrow (m+a)y = e^{mx}+ c(m+a)e^{-ax}$

$ \Rightarrow (m+a)y = e^{mx}+ Ce^{-ax} (where \quad C= c(m+a))$

Hence the required solution is $ (m+a)y = e^{mx}+ Ce^{-ax}$

Solution

Given that: $\frac{d y}{d x}+1=e^{x+y}$

Put $x+y=t$

Differentiate both sides w.r.t. x , (we get)

$ \begin{matrix} \therefore & 1+\frac{d y}{d x} & =\frac{d t}{d x} \\ \therefore & \frac{d t}{d x} & =e^{t} \Rightarrow \frac{d t}{e^{t}}=d x \\ \Rightarrow e^{-t} d t=d x \end{matrix} $

Integrating both sides, we have

$ \begin{gathered} \int e^{-t} d t=\int d x \Rightarrow-e^{-t}=x+c \\ \Rightarrow \quad-e^{-(x+y)}=x+c \Rightarrow \frac{-1}{e^{x+y}}=x+c \Rightarrow(x+c) e^{x+y}=-1 \end{gathered} $

Hence, the required solution is $(x+c) \ e^{x+y} \ +1=0$.

Solution

Given equation is $y d x-x d y=x^{2} y d x$.

$ \begin{aligned} & \Rightarrow \quad y d x-x^{2} y d x=x d y \\ & \Rightarrow \quad y(1-x^{2}) d x=x d y \\ & \Rightarrow \quad(\frac{1-x^{2}}{x}) d x=\frac{d y}{y} \Rightarrow(\frac{1}{x}-x) d x=\frac{d y}{y} \end{aligned} $

Integrating both sides we get

$ \begin{aligned} & \int(\frac{1}{x}-x) d x=\int \frac{d y}{y} \\ \Rightarrow & \log |x|-\frac{x^{2}}{2}=\log |y|+\log |c| \end{aligned} $

$ \begin{aligned} & \Rightarrow \log |x|-\frac{x^{2}}{2}=\log |y c| \Rightarrow \log |x|-\log |y c|=\frac{x^{2}}{2} \\ & \Rightarrow \log |\frac{x}{y c}|=\frac{x^{2}}{2} \Rightarrow |\frac{x}{y c}|=e^\frac{x^{2}}{2} \\ & \Rightarrow \quad \frac{x}{y c}=\pm e^{x^{2} / 2} \Rightarrow \frac{y c}{x}=\mp e^{-x^{2} / 2} \Rightarrow y c=\mp x e^{-x^{2} / 2} \\ & \therefore \quad y=\mp \frac{1}{c} \cdot x e^{-x^{2} / 2} \Rightarrow y=k x e^{-x^{2} / 2} \quad[where \quad k=\mp \frac{1}{c}] \end{aligned} $

Hence, the required solution is $y=k x e^{-x^{2} / 2}$.

Solution

Given equation is

$ \begin{aligned} \frac{d y}{d x} & =1+x+y^{2}+x y^{2} \\ \Rightarrow \quad \frac{d y}{d x} & =1(1+x)+y^{2}(1+x) \\ \Rightarrow \quad \frac{d y}{d x} & =(1+x)(1+y^{2})\\ \Rightarrow \quad \frac{d y}{1+y^{2}}=(1+x) d x \end{aligned} $

Integrating both sides, we get

$ \int \frac{d y}{1+y^{2}}=\int(1+x) d x \Rightarrow \tan ^{-1} y=x+\frac{x^{2}}{2}+c $

Put $x=0$ and $y=0$, we get $\tan ^{-1}(0)=0+0+c \Rightarrow c=0$

$\therefore \quad \tan ^{-1} y=x+\frac{x^{2}}{2} \Rightarrow y=\tan (x+\frac{x^{2}}{2})$

Hence, the required solution is $y=\tan (x+\frac{x^{2}}{2})$.

Solution

Given equation is $(x+2 y^{3}) \frac{d y}{d x}=y$

$ \begin{aligned} & \Rightarrow \quad \frac{d y}{d x}=\frac{y}{x+2 y^{3}} \Rightarrow \frac{d x}{d y}=\frac{x+2 y^{3}}{y} \\ & \Rightarrow \quad \frac{d x}{d y}=\frac{x}{y}+\frac{2 y^{3}}{y} \Rightarrow \frac{d x}{d y}-\frac{x}{y}=2 y^{2} \end{aligned} $

It is a linear differential equation of first order and first degree.

Here $P=-\frac{1}{y}$ and $Q=2 y^{2}$.

$\therefore$ Integrating factor I.F. $=e^{\int P d y}=e^{\int-\frac{1}{y} d y}=e^{-\log |y|}=e^{\log \frac{1}{|y|}}=\frac{1}{|y|}, \quad (y \neq 0)$

For $ y > 0 $

So the solution of the equation is

$ \begin{aligned} x \times I . F . & =\int Q \times \text{ I.F. } d y+c \\ x \cdot \frac{1}{y} & =\int 2 y^{2} \cdot \frac{1}{y} d y+c \\ \Rightarrow \quad \frac{x}{y} & =2 \int y d y+c \Rightarrow \frac{x}{y}=2 \cdot \frac{y^{2}}{2}+c \Rightarrow \frac{x}{y}=y^{2}+c \end{aligned} $

So $x=y(y^{2}+c)$

For $ y < 0 $

So the solution of the equation is

$ \begin{aligned} x \times I . F . & =\int Q \times \text{ I.F. } d y+k \\ x \cdot \frac{1}{-y} & =\int 2 y^{2} \cdot \frac{1}{-y} d y+k \\ \Rightarrow \quad \frac{x}{y} & =2 \int y d y-k \Rightarrow \frac{x}{y}=2 \cdot \frac{y^{2}}{2}-k \Rightarrow \frac{x}{y}=y^{2}-k \end{aligned} $

So $x=y(y^{2}-k)$

Hence, the required solution is $x=y(y^{2}+C)$.

Solution

Given equation is

$ \begin{aligned} & (\frac{2+\sin x}{1+y}) \frac{d y}{d x}=-\cos x \\ \Rightarrow &(\frac{2+\sin x}{\cos x}) \frac{d y}{d x}=-(1+y)\\ \Rightarrow \frac{d y}{(1+y)}=-(\frac{\cos x}{2+\sin x}) d x \end{aligned} $

Integrating both sides, we get

$ \begin{aligned} \int \frac{d y}{1+y} & =-\int \frac{\cos x}{2+\sin x} d x \\ \log |1+y| & =-\log |2+\sin x|+\log |c| \\ \Rightarrow \log |1+y|+\log |2+\sin x| & =\log |c| \\ \Rightarrow \quad \log |(1+y)(2+\sin x)| & =\log |c| \Rightarrow |(1+y)(2+\sin x)|=|c| \\ \Rightarrow (1+y)(2+\sin x)=\pm c \\ \Rightarrow (1+y)(2+\sin x)= C, \quad (where \quad C =\pm c) \end{aligned} $

Put $x=0$ and $y=1$, we get

$(1+1)(2+\sin 0)=C \Rightarrow 4=C$

$\therefore$ equation is $(1+y)(2+\sin x)=4$

Now put $x=\frac{\pi}{2}$

$ \begin{aligned} & \therefore \quad(1+y)(2+\sin \frac{\pi}{2})=4 \\ & \Rightarrow \quad(1+y)(2+1)=4 \Rightarrow 1+y=\frac{4}{3} \Rightarrow y=\frac{4}{3}-1 \Rightarrow y=\frac{1}{3} \end{aligned} $

So, $y(\frac{\pi}{2})=\frac{1}{3}$

Hence, the required solution is $y(\frac{\pi}{2})=\frac{1}{3}$.

Solution

Given equation is

$ (1+t) \frac{d y}{d t}-t y=1 \Rightarrow \frac{d y}{d t}-(\frac{t}{1+t}) y=\frac{1}{1+t} $

It is a linear differential equation of first order and first degree.

Here, $P=\frac{-t}{1+t}$ and $Q=\frac{1}{1+t}$

$\therefore$ Integrating factor I.F. $=e^{\int P d t}=e^{\int \frac{-t}{1+t} d t}=e^{-\int \frac{1+t-1}{1+t} d t}$

$ \begin{aligned} & I.F. =e^{-\int(1-\frac{1}{1+t}) d t}=e^{-[t-\log |(1+t)|]} \\ & =e^{-t+\log |(1+t)|}=e^{-t} \cdot e^{\log (1+t)} , \quad ((1+t)>0) \end{aligned} $

$\therefore$ I.F. $=e^{-t} \cdot(1+t)$

Required solution of the given differential equation is

$ \begin{matrix} y \times \text{ I.F. } =\int \text Q \times {I.F. } d t+c \\ \Rightarrow \qquad y \cdot e^{-t}(1+t) =\int \frac{1}{(1+t)} \cdot e^{-t} \cdot(1+t) d t+c \\ \Rightarrow \qquad y \cdot e^{-t}(1+t) =\int e^{-t} d t+c \\ \Rightarrow \qquad y \cdot e^{-t}(1+t) =-e^{-t}+c \end{matrix} $

$ \begin{aligned} & \text{ Put } t=0 \text{ and } y=-1 , [\because y(0)=-1] \\ & \Rightarrow \quad-1 \cdot e^{0} \cdot 1=-e^{0}+c \\ & \Rightarrow \quad-1=-1+c \Rightarrow c=0 \end{aligned} $

So the equation becomes

$ y e^{-t}(1+t)=-e^{-t} $

Now put $t=1$

$\therefore \quad y \cdot e^{-1}(1+1)=-e^{-1}$

$\Rightarrow \quad 2 y=-1 \Rightarrow y=-\frac{1}{2}$

Hence $y(1)=-\frac{1}{2}$ is verified.

Solution

Given equation is $y=(\sin ^{-1} x)^{2}+A \cos ^{-1} x+B$

$ \frac{d y}{d x}=2 \sin ^{-1} x \cdot \frac{1}{\sqrt{1-x^{2}}}+A \cdot(\frac{-1}{\sqrt{1-x^{2}}}) $

Multiplying both sides by $\sqrt{1-x^{2}}$, we get

$ \sqrt{1-x^{2}} \frac{d y}{d x}=2 \sin ^{-1} x-A $

Again differentiating w.r.t $x$, we get

$ \begin{aligned} & \sqrt{1-x^{2}} \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x} \cdot \frac{1 \times(-2 x)}{2 \sqrt{1-x^{2}}}=\frac{2}{\sqrt{1-x^{2}}} \\ \Rightarrow \quad & \sqrt{1-x^{2}} \frac{d^{2} y}{d x^{2}}-\frac{x}{\sqrt{1-x^{2}}} \frac{d y}{d x}=\frac{2}{\sqrt{1-x^{2}}} \end{aligned} $

Multiplying both sides by $\sqrt{1-x^{2}}$, we get

$ \Rightarrow \quad(1-x^{2}) \frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}-2=0 $

Which is the required differential equation.

Solution

Equation of circle which passes through the origin and whose centre lies on $y$-axis is

$$(x-0)^{2}+(y-a)^{2} =a^{2}$$

$$\Rightarrow \qquad x^{2}+y^{2}+a^{2}-2 a y =a^{2}$$

$$\Rightarrow \qquad x^{2}+y^{2}-2 a y =0 \tag{i}$$

Differentiate w.r.t. x , we get

$\Rightarrow \quad 2 x+2 y \cdot \frac{d y}{d x}-2 a \cdot \frac{d y}{d x}=0$

$\Rightarrow \quad x+y \frac{d y}{d x}-a \cdot \frac{d y}{d x}=0 \Rightarrow x+(y-a) \cdot \frac{d y}{d x}=0$

$ y-a=\frac{-x}{d y / d x} $

$ \therefore \quad a=y+\frac{x}{d y / d x} $

Putting the value of $a$ in eq. (i), we get

$ \begin{aligned} & x^{2}+y^{2}-2(y+\frac{x}{d y / d x}) y=0 \\ & \Rightarrow \quad x^{2}+y^{2}-2 y^{2}-\frac{2 x y}{\frac{d y}{d x}}=0 \Rightarrow x^{2}-y^{2}=\frac{2 x y}{\frac{d y}{d x}} \\ & \therefore \quad(x^{2}-y^{2}) \frac{d y}{d x}-2 x y=0 \end{aligned} $

Hence, the required differential equation is

$ (x^{2}-y^{2}) \frac{d y}{d x}-2 x y=0 $

Solution

Given equation is

$ \begin{aligned} & (1+x^{2}) \frac{d y}{d x}+2 x y=4 x^{2} \\ & \Rightarrow \quad \frac{d y}{d x}+\frac{2 x}{1+x^{2}} \cdot y=\frac{4 x^{2}}{1+x^{2}} \end{aligned} $

It is a linear differential equation of first order and first degree.

Here, $P=\frac{2 x}{1+x^{2}}$ and $Q=\frac{4 x^{2}}{1+x^{2}}$

Integrating factor I.F. $=e^{\int P d x}=e^{\int \frac{2 x}{1+x^{2}} d x}=e^{\log |(1+x^{2})|}=|(1+x^{2})| = (1+ x^{2} ) , \quad (\because \quad (1+ x^{2} ) \quad is \quad positive) $

$\therefore$ Solution is

$$ \begin{align*} & y \times \text{ I.F. }=\int Q \times \text{ I.F. } d x+c \\ & \Rightarrow y(1+x^{2})=\int \frac{4 x^{2}}{1+x^{2}} \times(1+x^{2}) d x+c \\ & \Rightarrow \quad y(1+x^{2})=\int 4 x^{2} d x+c \\ & \Rightarrow \quad y(1+x^{2})=\frac{4}{3} x^{3}+c \tag{i} \end{align*} $$

Since the curve is passing through origin i.e., $(0,0)$

$\therefore$ Put $y=0$ and $x=0$ in eq. (i)

$0(1+0)=\frac{4}{3}(0)^{3}+c \Rightarrow c=0$

$\therefore$ Equation is $y(1+x^{2})=\frac{4}{3} x^{3} \Rightarrow y=\frac{4 x^{3}}{3(1+x^{2})}$

Hence, the required solution is $y=\frac{4 x^{3}}{3(1+x^{2})}$.

Solution

Given equation is $x^{2} \frac{d y}{d x}=x^{2}+x y+y^{2}$

$ \Rightarrow \quad \frac{d y}{d x}=\frac{x^{2}+x y+y^{2}}{x^{2}} $ (Homogenious differential equation )

Put $y=v x$

$\therefore \quad \frac{d y}{d x}=v+x \cdot \frac{d v}{d x}$

$\therefore \quad v+x \cdot \frac{d v}{d x}=\frac{x^{2}+v x^{2}+v^{2} x^{2}}{x^{2}}$

$\Rightarrow \quad v+x \cdot \frac{d v}{d x}=\frac{x^{2}(1+v+v^{2})}{x^{2}}$

$\Rightarrow \quad v+x \cdot \frac{d v}{d x}=1+v+v^{2} \Rightarrow x \cdot \frac{d v}{d x}=1+v+v^{2}-v$

$ \Rightarrow \quad x \cdot \frac{d v}{d x}=1+v^{2}\\ \Rightarrow \frac{d v}{1+v^{2}}=\frac{d x}{x} $

Integrating both sides, we get

$ \begin{aligned} \int \frac{d v}{1+v^{2}} & =\int \frac{d x}{x} \\ \Rightarrow \quad \tan ^{-1} v & =\log |x|+c \Rightarrow \tan ^{-1}(\frac{y}{x})=\log |x|+c \end{aligned} $

Hence, the required solution is $\tan ^{-1}(\frac{y}{x})=\log |x|+c$.

Solution

Given equation is

$ \begin{aligned} & (1+y^{2})+(x-e^{\tan ^{-1} y}) \frac{d y}{d x}=0 \\ \Rightarrow & (x-e^{\tan ^{-1} y}) \frac{d y}{d x}=-(1+y^{2}) \Rightarrow \frac{d y}{d x}=\frac{-(1+y^{2})}{x-e^{\tan ^{-1} y}} \\ \Rightarrow & \quad \frac{d x}{d y}=\frac{x-e^{\tan ^{-1} y}}{-(1+y^{2})} \Rightarrow \frac{d x}{d y}=-\frac{x}{(1+y^{2})}+\frac{e^{\tan ^{-1} y}}{1+y^{2}} \\ \Rightarrow & \frac{d x}{d y}+\frac{x}{(1+y^{2})}=\frac{e^{\tan ^{-1} y}}{1+y^{2}} \end{aligned} $

It is a linear differential equation of first order and first degree.

Here, $P=\frac{1}{1+y^{2}}$ and $Q=\frac{e^{\tan ^{-1} y}}{1+y^{2}}$

$\therefore$ Integrating factor I.F. $=e^{\int P d y}=e^{\int \frac{1}{1+y^{2}} d y}=e^{\tan ^{-1} y}$

$\therefore$ Solution is

$ \begin{aligned} & x \times \text{ I.F. }=\int Q \times \text{ I.F. } d y+c \\ \Rightarrow \quad x \cdot e^{\tan ^{-1} y} & =\int \frac{e^{\tan ^{-1} y}}{1+y^{2}} \cdot e^{\tan ^{-1} y} d y+c \end{aligned} $

Put $e^{\tan ^{-1} y}=t$

$\therefore e^{\tan ^{-1} y} \cdot \frac{1}{1+y^{2}} d y=d t$

$\therefore \quad x \cdot e^{\tan ^{-1} y}=\int t \cdot d t+c$

$\Rightarrow x \cdot e^{\tan ^{-1} y}=\frac{1}{2} t^{2}+c$

$ \begin{aligned} & \Rightarrow \quad x \cdot e^{\tan ^{-1} y}=\frac{1}{2}(e^{\tan ^{-1} y})^{2}+c \\ & \Rightarrow 2 x \cdot e^{\tan ^{-1} y}=(e^{\tan ^{-1} y})^{2}+2 c \\ & \Rightarrow 2 x \cdot e^{\tan ^{-1} y}=(e^{2\tan ^{-1} y})+ C, \quad (where \quad C = 2c) \end{aligned} $

Hence, this is the required general solution.

Solution

The given equation is $y^{2} d x+(x^{2}-x y+y^{2}) d y=0$.

$ \begin{matrix} \Rightarrow & y^{2} d x=-(x^{2}-x y+y^{2}) d y \\ \Rightarrow & \frac{d x}{d y}=-\frac{x^{2}-x y+y^{2}}{y^{2}} \end{matrix} $

Since it is a homogeneous differential equation

$\therefore$ Put $x=v y \Rightarrow \frac{d x}{d y}=v+y \cdot \frac{d v}{d y}$

So, $\quad v+y \cdot \frac{d v}{d y}=-(\frac{v^{2} y^{2}-v y^{2}+y^{2}}{y^{2}})$

$\Rightarrow \quad v+y \cdot \frac{d v}{d y}=-\frac{y^{2}(v^{2}-v+1)}{y^{2}}$

$\Rightarrow \quad v+y \cdot \frac{d v}{d y}=(-v^{2}+v-1) \Rightarrow y \cdot \frac{d v}{d y}=-v^{2}+v-1-v$

$\Rightarrow \quad y \cdot \frac{d v}{d y}=-v^{2}-1 \Rightarrow \frac{d v}{(v^{2}+1)}=-\frac{d y}{y}$

Integrating both sides, we get

$ \begin{aligned} & \Rightarrow \quad \int \frac{d v}{(v^{2}+1)}=-\int \frac{d y}{y} \Rightarrow \tan ^{-1} v=-\log |y|+c \\ & \Rightarrow \quad \tan ^{-1}(\frac{x}{y})+\log |y|=c \end{aligned} $

Hence the required solution is $\tan ^{-1}(\frac{x}{y})+\log |y|=c$.

Solution

Given differential equation is

$\qquad (x+y)(d x-d y) =d x+d y$

$\Rightarrow \quad (x+y) d x-(x+y) d y =d x+d y$

$\Rightarrow \quad -(x+y) d y-d y =d x-(x+y) d x$

$ \begin{aligned} \Rightarrow \quad -(x+y+1) d y =-(x+y-1) d x \\ \Rightarrow \frac{d y}{d x} =\frac{x+y-1}{x+y+1} \end{aligned} $

Put $x+y=z$

$ \begin{aligned} \therefore \quad 1+\frac{d y}{d x} & =\frac{d z}{d x} \\ \frac{d y}{d x} & =\frac{d z}{d x}-1 \end{aligned} $

So, $\quad \frac{d z}{d x}-1=\frac{z-1}{z+1}$

$\Rightarrow \quad \frac{d z}{d x}=\frac{z-1}{z+1}+1 \Rightarrow \frac{d z}{d x}=\frac{z-1+z+1}{z+1}$

$\Rightarrow \quad \frac{d z}{d x}=\frac{2 z}{z+1} \quad \Rightarrow \frac{z+1}{z} d z=2 \cdot d x$

Integrating both sides, we get

$ \begin{aligned} & \int \frac{z+1}{z} d z=2 \int d x \\ & \Rightarrow \quad \int(1+\frac{1}{z}) d z=2 \int d x \\ & \Rightarrow \quad z+\log |z|=2 x+\log |c| \\ & \Rightarrow \quad x+y+\log |x+y|=2 x+\log |c| \\ & \Rightarrow \quad y+\log |x+y|=x+\log |c| \\ & \Rightarrow \quad \log |x+y|=x-y+\log |c| \\ & \Rightarrow \log |x+y|-\log |c|=(x-y) \\ & \Rightarrow \quad \log |\frac{x+y}{c}|=(x-y) \Rightarrow |\frac{x+y}{c}|=e^{x-y} \\ & \quad x+y=\pm c \cdot e^{x-y} \\ & \therefore \quad x+y=C \cdot e^{x-y}, \quad (where \quad C = \pm c) \end{aligned} $

Hence, the required solution is $x+y=C \cdot e^{x-y}$.

Solution

Given differential equation is

$ \begin{aligned} 2(y+3)-x y \cdot \frac{d y}{d x} & =0 \\ \Rightarrow \quad x y \cdot \frac{d y}{d x} & =2 y+6 \\ \Rightarrow \quad(\frac{y}{2 y+6}) d y & =\frac{d x}{x} \Rightarrow \frac{1}{2}(\frac{y}{y+3}) d y=\frac{d x}{x} \end{aligned} $

Integrating both sides, we get

$ \begin{aligned} & \Rightarrow \quad \frac{1}{2} \int \frac{y}{y+3} \cdot d y=\int \frac{d x}{x} \Rightarrow \frac{1}{2} \int \frac{y+3-3}{y+3} d y=\int \frac{d x}{x} \\ & \Rightarrow \quad \frac{1}{2} \int(1-\frac{3}{y+3}) d y=\int \frac{d x}{x} \end{aligned} $

$ \begin{aligned} & \Rightarrow \frac{1}{2} \int 1 \cdot d y-\frac{3}{2} \int \frac{1}{y+3} d y=\int \frac{d x}{x} \\ & \Rightarrow \quad \frac{1}{2} y-\frac{3}{2} \log |y+3|=\log|x|+\log |c| \\ & \Rightarrow \quad \frac{1}{2} y-\frac{3}{2} \log |y+3|=\log|xc| \\ & \Rightarrow \quad y-{3} \log |y+3|=2\log|xc| \\ & \Rightarrow \quad y-{3} \log |y+3|=\log(|xc|)^{2} \\ & \Rightarrow \quad y= \log (|y+3|)^{3}+\log(|xc|)^{2} \\ & \Rightarrow \quad y= \log (|y+3|)^{3}+\log(xc)^{2} \\ & \Rightarrow \quad y= \log ((xc)^{2}(|y+3|)^{3}) \\ & \Rightarrow \quad e^{y}= ((xc)^{2}(|y+3|)^{3}) \\ & \Rightarrow \quad x^{2}|y+3|^{3}=\frac {1}{c^{2}} e^{y} \\ & \Rightarrow \quad x^{2}(y+3)^{3}=\pm \frac {1}{c^{2}} e^{y} \\ & \Rightarrow \quad x^{2}(y+3)^{3}=C e^{y},(where \quad C=\pm \frac {1}{c^{2}}) \end{aligned} $

Put $x=1, y=-2$

$ \Rightarrow \quad 1^{2}(-2+3)^{3} =C e^{-2} $

$ \Rightarrow \quad C=e^{2}$

$ \therefore$ equation is

$ \Rightarrow x^{2}(y+3)^{3}=e^{y+2}$

Hence, the required solution is $x^{2}(y+3)^{3}=e^{y+2}$.

Solution

The given differential equation is

$ d y=\cos x \text{(2-y cosec x)} d x $

$\Rightarrow \frac{d y}{d x}=\cos x\text {(2-y cosec x)} \Rightarrow \frac{d y}{d x}=2 \cos x-y \cos x \cdot \frac{1}{sinx}$

$\Rightarrow \frac{d y}{d x}=2 \cos x-y \cot x \Rightarrow \frac{d y}{d x}+y \cot x=2 \cos x$

It is a linear differential equation of first order and first degree.

Here, $P=\cot x$ and $Q=2 \cos x$.

$\therefore$ Integrating factor I.F. $=e^{\int P d x}=e^{\int \cot x d x}=e^{\log |\sin x|}=\sin x , \quad (\sin x>0)$

$\therefore$ Required solution is $y \times$ I.F $=\int Q \times$ I.F. $d x+c$ $\Rightarrow y \cdot \sin x=\int 2 \cos x \cdot \sin x d x+c$

$\Rightarrow y \cdot \sin x=\int \sin 2 x d x+c \Rightarrow y \cdot \sin x=-\frac{1}{2} \cos 2 x+c$

Put $x=\frac{\pi}{2}$ and $y=2$, we get

$ \begin{aligned} 2 \sin \frac{\pi}{2} & =-\frac{1}{2} \cos \pi+c \\ \Rightarrow \quad 2(1) & =-\frac{1}{2}(-1)+c \Rightarrow 2=\frac{1}{2}+c \Rightarrow c=2-\frac{1}{2}=\frac{3}{2} \end{aligned} $

$\therefore$ The equation is $y \sin x=-\frac{1}{2} \cos 2 x+\frac{3}{2}$.

Solution

Given that $A x^{2}+B y^{2}=1$

Differentiating w.r.t. $x$, we get

$\qquad 2 A \cdot x+2 B y \frac{d y}{d x} =0$

$\Rightarrow \qquad A x+B y \cdot \frac{d y}{d x} =0 \Rightarrow B y \cdot \frac{d y}{d x}=-A x$

$\therefore \qquad \frac{y}{x} \cdot \frac{d y}{d x} =-\frac{A}{B}$

Differentiating both sides again w.r.t. $x$, we have

$ \begin{aligned} & \frac{y}{x} \cdot \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}(\frac{x \cdot \frac{d y}{d x}-y \cdot 1}{x^{2}})=0 \\ \Rightarrow & \frac{y x^{2}}{x} \cdot \frac{d^{2} y}{d x^{2}}+x \cdot(\frac{d y}{d x})^{2}-y \cdot \frac{d y}{d x}=0 \\ \Rightarrow & x y \cdot \frac{d^{2} y}{d x^{2}}+x \cdot(\frac{d y}{d x})^{2}-y \cdot \frac{d y}{d x}=0 \Rightarrow x y \cdot y^{\prime \prime}+x \cdot(y^{\prime})^{2}-y \cdot y^{\prime}=0 \end{aligned} $

Hence, the required equation is

$ x y \cdot y^{\prime \prime}+x \cdot(y^{\prime})^{2}-y \cdot y^{\prime}=0 $

Solution

Given differential equation is

$ \begin{matrix} (1+y^{2}) \tan ^{-1} x d x+2 y(1+x^{2}) d y=0 \\ \Rightarrow \qquad 2 y(1+x^{2}) d y=-(1+y^{2}) \cdot \tan ^{-1} x \cdot d x \\ \Rightarrow \qquad \frac{2 y}{1+y^{2}} d y=-\frac{\tan ^{-1} x}{1+x^{2}} \cdot d x \end{matrix} $

Integrating both sides, we get

$\int \frac{2 y}{1+y^{2}} d y =-\int \frac{\tan ^{-1} x}{1+x^{2}} \cdot d x $

$ let \quad 1+y^{2} = t $

$ 2y dy = dt $

$ and \quad \tan ^{-1} x = k $

$ \frac{1} {1+x^{2}} dx = dk $

$\therefore \int \frac {dt}{t} = -\int kdk $

$ log |t| =- \frac {k^{2}}{2} + c $

$\Rightarrow \quad \log |1+y^{2}| =-\frac{1}{2}(\tan ^{-1} x)^{2}+c$

$\Rightarrow \quad \frac{1}{2}(\tan ^{-1} x)^{2}+\log |1+y^{2}|=c$

$\Rightarrow \quad \frac{1}{2}(\tan ^{-1} x)^{2}+\log (1+y^{2})=c, \quad (\because (1+y)^{2} \quad is \quad positive)$

Which is the required solution.

Solution

Family of concentric circles with centre $(1,2)$ and radius ’ $r$ ’ is

$ (x-1)^{2}+(y-2)^{2}=r^{2} $

Differentiating both sides w.r.t., $x$ we get

$ 2(x-1)+2(y-2) \frac{d y}{d x}=0 \Rightarrow(x-1)+(y-2) \frac{d y}{d x}=0 $

Which is the required equation.

Solution

The given differential equation is

$ \begin{aligned} & y+\frac{d}{d x}(x y)=x(\sin x+\log x) \\ & \Rightarrow \quad y+x \cdot \frac{d y}{d x}+y=x(\sin x+\log x) \\ & \Rightarrow \quad x \frac{d y}{d x}=x(\sin x+\log x)-2 y \\ & \Rightarrow \quad \frac{d y}{d x}=(\sin x+\log x)-\frac{2 y}{x} \Rightarrow \frac{d y}{d x}+\frac{2}{x} y=(\sin x+\log x) \end{aligned} $

It is a linear differential equation of first order and first degree.

Here, $P=\frac{2}{x}$ and $Q=(\sin x+\log x)$

Integrating factor I.F. $=e^{\int P d x}=e^{\int \frac{2}{x} d x}=e^{2 \log |x|}=e^{\log |x|^{2}}=|x|^{2}= x^{2}$

$\therefore$ Solution is

$$ \begin{align*} y \times \text{ I.F. } & =\int Q \times \text{ I.F. } d x+c \\ \Rightarrow \quad y \cdot x^{2} & =\int(\sin x+\log x) x^{2} d x+c \tag{1}\\ \Rightarrow \quad y \cdot x^{2} & = \text {I +c } \end{align*} $$

Let I

$ \begin{aligned} & =\int(\sin x+\log x) x^{2} d x \\ & =\int x^{2} \sin x d x+\int x^{2} \log x d x \\ & (Integration \quad by \quad parts) \\ & \begin{aligned} = & {[x^{2} \cdot \int \sin x d x-\int(D(x^{2}) \cdot \int \sin x d x) d x]+} {[\log x \cdot \int x^{2} d x-\int(D(\log x) \cdot \int x^{2} d x) d x] } \end{aligned} \\ & =[x^{2}(-\cos x)-2 \int-x \cos x d x]+[\log x \cdot \frac{x^{3}}{3}-\int \frac{1}{x} \cdot \frac{x^{3}}{3} d x] \\ & =[-x^{2} \cos x+2(x \sin x-\int 1 \cdot \sin x d x)]+[\frac{x^{3}}{3} \log x-\frac{1}{3} \int x^{2} d x] \\ & =-x^{2} \cos x+2 x \sin x+2 \cos x+\frac{x^{3}}{3} \log x-\frac{1}{9} x^{3} \end{aligned} $

Now from eq (1) we get,

$ \begin{aligned} y \cdot x^{2} & =-x^{2} \cos x+2 x \sin x+2 \cos x+\frac{x^{3}}{3} \log x-\frac{1}{9} x^{3}+c \\ \therefore \quad & y=-\cos x+\frac{2 \sin x}{x}+\frac{2 \cos x}{x^{2}}+\frac{x \log x}{3}-\frac{1}{9} x+c \cdot x^{-2} \end{aligned} $

Hence, the required solution is

$ y=-\cos x+\frac{2 \sin x}{x}+\frac{2 \cos x}{x^{2}}+\frac{x \log x}{3}-\frac{1}{9} x+c \cdot x^{-2} $

Solution

Given that: $(1+\tan y)(d x-d y)+2 x d y=0$

$\Rightarrow \quad(1+\tan y) d x-(1+\tan y) d y+2 x d y=0$

$ \Rightarrow \quad(1+\tan y) d x-(1+\tan y-2 x) d y=0 $

$ \begin{aligned} & \Rightarrow \quad(1+\tan y) \frac{d x}{d y}=(1+\tan y-2 x) \Rightarrow \frac{d x}{d y}=\frac{1+\tan y-2 x}{1+\tan y} \\ & \Rightarrow \quad \frac{d x}{d y}=1-\frac{2 x}{1+\tan y} \Rightarrow \frac{d x}{d y}+\frac{2 x}{1+\tan y}=1 \end{aligned} $

It is a linear differential equation of first order and first degree.

Here, $P=\frac{2}{1+\tan y}$ and $Q=1$

Integrating factor I.F.

$ \begin{aligned} & =e^{\int \frac{2}{1+\tan y} d y}=e^{\int \frac{2 \cos . y}{\sin y+\cos y} d y} \\ & =e^{\int \frac{\sin y+\cos y-\sin y+\cos y}{(\sin y+\cos y)} d y}=e^{\int(1+\frac{\cos y-\sin y}{\sin y+\cos y}) d y} \\ & =e^{\int 1 \cdot d y} \cdot e^{\int \frac{\cos y-\sin y}{\sin y+\cos y} d y} \\ & =e^{y} \cdot e^{\log |(\sin y+\cos y)|}=e^{y} \cdot(\sin y+\cos y), \quad ((\sin y+\cos y)>0) \end{aligned} $

So, the solution is $\quad x \times$ I.F. $=\int Q \times$ I.F. $d y+c$

$\Rightarrow x \cdot e^{y}(\sin y+\cos y)=\int 1 \cdot e^{y}(\sin y+\cos y) d y+c$

$\Rightarrow \quad x \cdot e^{y}(\sin y+\cos y)=e^{y} \cdot \sin y+c$

$ [\because \int e^{x}[f(x)+f^{\prime}(x)] d x=e^{x} f(x)+c] $

$\Rightarrow \quad x(\sin y+\cos y)=\sin y+c \cdot e^{-y}$

Hence, the required solution is $x(\sin y+\cos y)=\sin y+c \cdot e^{-y}$.

Solution

Given that : $\frac{d y}{d x}=\cos (x+y)+\sin (x+y)$

Put $\quad x+y=v$, on differentiating w.r.t. $x$, we get,

$1+\frac{d y}{d x} =\frac{d v}{d x}$

$\therefore \quad \frac{d y}{d x} =\frac{d v}{d x}-1$

$\therefore \quad \frac{d v}{d x}-1 =\cos v+\sin v$

$\Rightarrow \quad \frac{d v}{d x} =\cos v+\sin v+1$

$\Rightarrow \quad \frac{d v}{\cos v+\sin v+1} = d x $

$\Rightarrow \quad \frac{d v}{(\frac{1-\tan ^{2} \frac{v}{2}}{1+\tan ^{2} \frac{v}{2}}+\frac{2 \tan \frac{v}{2}}{1+\tan ^{2} \frac{v}{2}}+1)} = d x $

$\Rightarrow \quad \frac{(1+\tan ^{2} \frac{v}{2})}{1-\tan ^{2} \frac{v}{2}+2 \tan \frac{v}{2}+1+\tan ^{2} \frac{v}{2}} d v = d x $

Integrating \quad both \quad sides, we \quad have $

$\Rightarrow \quad \int \frac{\sec ^{2} \frac{v}{2}}{2+2 \tan \frac {v}{2}} d v =\int 1 . d x $

$\text{ Put } \quad 2+2 \tan \frac{v}{2} = t$

$2 \cdot \frac{1}{2} \sec ^{2} \frac{v}{2} d v =d t \Rightarrow \sec ^{2} \frac{v}{2} d v=d t $

$\int \frac{d t}{t} =\int 1 \cdot d x $

$\Rightarrow \quad \log |t| =x+c $

$\Rightarrow \quad \log |2+2 \tan \frac{v}{2}| =x+c $

$\Rightarrow \quad \log |2+2 \tan (\frac{x+y}{2})|=x+c \Rightarrow \log 2(|1+\tan (\frac{x+y}{2})|)=x+c $

$\Rightarrow \quad \log 2+\log (|1+\tan (\frac{x+y}{2})|) =x+c $

$\Rightarrow \quad \log (|1+\tan (\frac{x+y}{2})|) =x+c-\log 2$

Hence, the required solution is

$ \log (|1+\tan (\frac{x+y}{2})|)=x+K \quad[c-\log 2=K] $

Solution

Given equation is $\frac{d y}{d x}-3 y=\sin 2 x$.

It is a linear differential equation of first order and first degree.

Here, $P=-3$ and $Q=\sin 2 x$

$\therefore$ Integrating factor I.F. $=e^{\int P d x}=e^{\int-3 d x}=e^{-3 x}$

$\therefore$ Solution is

$ \begin{aligned} & y \times \text{ I.F. }=\int Q \times \text{ I.F. } d x+c \\ & \Rightarrow \quad y \cdot e^{-3 x}=\int \sin 2 x \cdot e^{-3 x} d x+c \\ & \text{ Let } \quad I=\int \sin 2 x \cdot e^{-3 x} d x \\ Integration \quad by \quad part \\ & \Rightarrow \quad I=\sin 2 x \cdot \int e^{-3 x} d x-\int(D(\sin 2 x) \cdot \int e^{-3 x} d x) d x \\ & \Rightarrow \quad I=\sin 2 x \cdot \frac{e^{-3 x}}{-3}-\int 2 \cos 2 x \cdot \frac{e^{-3 x}}{-3} d x \\ & \Rightarrow \quad I=\frac{e^{-3 x}}{-3} \sin 2 x+\frac{2}{3} \int {\cos 2 x} \cdot e^{-3 x} d x \\ & \Rightarrow \quad I=\frac{e^{-3 x}}{-3} \sin 2 x+\frac{2}{3}[\cos 2 x \cdot \int e^{-3 x} d x-\int[D (\cos 2 x) \cdot \int e^{-3 x} d x] d x] \end{aligned} $

$ \begin{aligned} \Rightarrow \quad I & =\frac{e^{-3 x}}{-3} \sin 2 x+\frac{2}{3}[\cos 2 x \cdot \frac{e^{-3 x}}{-3}- \int -2 \sin 2 x \cdot \frac{e^{-3 x}}{-3}] d x \\ \Rightarrow \quad I & =\frac{e^{-3 x}}{-3} \sin 2 x-\frac{2}{9} \cos 2 x \cdot e^{-3 x}-\frac{4}{9} \int \sin 2 x \cdot e^{-3 x} d x \\ \Rightarrow \quad & =\frac{e^{-3 x}}{-3} \sin 2 x-\frac{2}{9} e^{-3 x} \cos 2 x-\frac{4}{9} I \\ \Rightarrow \quad I+\frac{4}{9} I & =\frac{e^{-3 x}}{-3} \sin 2 x-\frac{2}{9} e^{-3 x} \cos 2 x \\ \Rightarrow \quad \frac{13 I}{9} & =-\frac{1}{9}[3 e^{-3 x} \sin 2 x+2 e^{-3 x} \cos 2 x] \\ \Rightarrow \quad I & =-\frac{1}{13} e^{-3 x}[3 \sin 2 x+2 \cos 2 x] \end{aligned} $

$\therefore$ The equation becomes

$ \begin{aligned} y \cdot e^{-3 x} & =-\frac{1}{13} e^{-3 x}[3 \sin 2 x+2 \cos 2 x]+c \\ \therefore \quad y & =-\frac{1}{13}[3 \sin 2 x+2 \cos 2 x]+c \cdot e^{3 x} \end{aligned} $

Hence, the required solution is

$ y=-[\frac{3 \sin 2 x+2 \cos 2 x}{13}]+c \cdot e^{3 x} $

Solution

Given that the slope of tangent to a curve at $(x, y)$ is

$ \frac{d y}{d x}=\frac{x^{2}+y^{2}}{2 x y} $

It is a homogeneous differential equation

So, put $y=v x \Rightarrow \frac{d y}{d x}=v+x \cdot \frac{d v}{d x}$

$ v+x \cdot \frac{d v}{d x}=\frac{x^{2}+v^{2} x^{2}}{2 x \cdot v x} $

$ \begin{aligned} & \Rightarrow \quad v+x \cdot \frac{d v}{d x}=\frac{1+v^{2}}{2 v} \\ & \Rightarrow \quad x \cdot \frac{d v}{d x}=\frac{1+v^{2}}{2 v}-v \Rightarrow x \cdot \frac{d v}{d x}=\frac{1+v^{2}-2 v^{2}}{2 v} \\ & \Rightarrow \quad x \cdot \frac{d v}{d x}=\frac{1-v^{2}}{2 v} \Rightarrow \frac{2 v}{1-v^{2}} d v=\frac{d x}{x} \end{aligned} $

Integrating both sides, we get

$ \begin{gathered} \int \frac{2 v}{1-v^{2}} d v=\int \frac{d x}{x} \Rightarrow-\log |1-v^{2}|=\log |x|+\log |C| \\ \Rightarrow-\log |1-\frac{y^{2}}{x^{2}}|=\log |x|+\log |C| \Rightarrow-\log |\frac{x^{2}-y^{2}}{x^{2}}|=\log |x|+\log |C| \\ \Rightarrow \log |\frac{x^{2}}{x^{2}-y^{2}}|=\log |x C| \Rightarrow \frac{x^{2}}{x^{2}-y^{2}}=\pm Cx \\ \Rightarrow \frac{x^{2}}{x^{2}-y^{2}}=cx, \quad (where \quad c = \pm C) \end{gathered} $

Since, the curve is passing through the point $(2,1)$

$ \therefore \quad \frac{(2)^{2}}{(2)^{2}-(1)^{2}}=2 c \Rightarrow \frac{4}{3}=2 c \Rightarrow c=\frac{2}{3} $

Hence, the required equation is

$ \frac{x^{2}}{x^{2}-y^{2}}=\frac{2}{3} x \Rightarrow 2(x^{2}-y^{2})=3 x $

Solution

Given that the slope of the tangent to the curve at $(x, y)$ is

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

Integrating both sides, we have

$ \begin{aligned} & \int \frac{d y}{y-1}=\int \frac{d x}{x^{2}+x} \\ & \Rightarrow \quad \int \frac{d y}{y-1}=\int \frac{d x}{x^{2}+x+\frac{1}{4}-\frac{1}{4}} \text{ [making perfect square] } \\ & \Rightarrow \quad \int \frac{d y}{y-1}=\int \frac{d x}{(x+\frac{1}{2})^{2}-(\frac{1}{2})^{2}} \end{aligned} $

$ \begin{matrix} \Rightarrow & \log |y-1|=\frac{1}{2 \times \frac{1}{2}} \log |\frac{x+\frac{1}{2}-\frac{1}{2}}{x+\frac{1}{2}+\frac{1}{2}}|+\log |c| \\ \Rightarrow & \log |y-1|=\log |\frac{x}{x+1}|+\log |c| \\ \Rightarrow & \log |y-1|=\log |c(\frac{x}{x+1})| \\ \therefore & |y-1|=|\frac{c x}{x+1}| \\ \therefore & y-1=\pm \frac{c x}{x+1} \\ \Rightarrow & (y-1)(x+1)=C x , \quad (where \quad C= \pm c) \end{matrix} $

Since, the line is passing through the point $(1,0)$, then $(0-1)(1+1)=C(1) \Rightarrow C=-2$.

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

Hence, the required solution is $(y-1)(x+1)+2 x=0$.

Solution

Here, slope of the tangent of the curve $=\frac{d y}{d x}$ and the difference between the abscissa and ordinate $=x-y$.

$\therefore$ As per the condition, $\frac{d y}{d x}=(x-y)^{2}$

Put $x-y=v$

$ \begin{aligned} & 1-\frac{d y}{d x}=\frac{d v}{d x} \\ & \therefore \quad \frac{d y}{d x}=1-\frac{d v}{d x} \end{aligned} $

$\therefore$ the equation becomes

$ 1-\frac{d v}{d x}=v^{2} \Rightarrow \frac{d v}{d x}=1-v^{2} \Rightarrow \frac{d v}{1-v^{2}}=d x $

Integrating both sides, we get

$$ \begin{align*} \int \frac{d v}{1-v^{2}} & =\int d x \\ \Rightarrow \quad \frac{1}{2} \log |\frac{1+v}{1-v}| & =x+c \quad \Rightarrow \quad \frac{1}{2} \log |\frac{1+x-y}{1-x+y}|=x+c \\ \Rightarrow \quad |\frac{1+x-y}{1-x+y}|=e^{2(x+c)} \\ \Rightarrow \quad |\frac{1+x-y}{1-x+y}|=e^{2c}e^{2x} \\ \Rightarrow \quad \frac{1+x-y}{1-x+y}=\pm e^{2c}e^{2x} \\ \Rightarrow \quad \frac{1+x-y}{1-x+y}=\pm e^{2c}e^{2x} \\ \Rightarrow \quad \frac{1+x-y}{1-x+y}=Ce^{2x} (where C= \pm e^{2c}) \end{align*} $$

Since, the curve is passing through $(0,0)$

$ \Rightarrow \frac{1+0-0}{1-0+0} =Ce^{ 0} $

$ \Rightarrow C=1 $

$ \begin{aligned} & \therefore & \frac{1+x-y}{1-x+y} & =e^{2 x} \\ \Rightarrow & & (1+x-y) & =e^{2 x}(1-x+y) \end{aligned} $

Hence, the required equation is $(1+x-y)=e^{2 x}(1-x+y)$.

Solution

Let $P(x, y)$ be any point on the curve and $AB$ be the tangent to the given curve at $P$.

$P$ is the mid point of $A B$ (given)

$\therefore$ Coordinates of $A$ and $B$ are $(2 x, 0)$ and $(0,2 y)$ respectively.

$\therefore$ Slope of the tangent

$ AB= $

$ \begin{aligned} & \frac{2 y-0}{0-2 x}=-\frac{y}{x} \\ & \therefore \quad \frac{d y}{d x}=-\frac{y}{x} \Rightarrow \frac{d y}{y}=-\frac{d x}{x} \end{aligned} $

Integrating both sides, we get

$ \begin{aligned} & \int \frac{d y}{y}=-\int \frac{d x}{x} \Rightarrow \log |y|=-\log |x|+\log |c| \\ & \Rightarrow \quad \log |y|+\log |x|=\log |c| \quad \Rightarrow \log |y x|=\log|c| \\ & \therefore \quad |y x|=|c| \\ & \therefore \quad y x=\pm c \\ & \therefore \quad y x=C , \quad (where \quad C = \pm c) \end{aligned} $

Since, the curve passes through $(1,1)$

$ \begin{aligned} \therefore & & 1 \times 1 & =C\quad \therefore \quad C=1 \\ \Rightarrow & & y x & =1 \end{aligned} $

Hence, the required equation is $x y=1$.

Solution

Given that: $x \frac{d y}{d x}=y(\log y-\log x+1)$

$ \Rightarrow \quad x \frac{d y}{d x}=y[\log (\frac{y}{x})+1] \Rightarrow \frac{d y}{d x}=\frac{y}{x}[\log (\frac{y}{x})+1] $

Since, it is a homogeneous differential equation.

$\therefore$ Put $y=v x \Rightarrow \frac{d y}{d x}=v+x \cdot \frac{d v}{d x}$

$\therefore \quad v+x \cdot \frac{d v}{d x}=\frac{v x}{x}[\log (\frac{v x}{x})+1]$

$\Rightarrow \quad v+x \cdot \frac{d v}{d x}=v[\log v+1]$

$\Rightarrow \quad x \cdot \frac{d v}{d x}=v[\log v+1]-v \Rightarrow x \cdot \frac{d v}{d x}=v[\log v+1-1]$

$\Rightarrow \quad x \cdot \frac{d v}{d x}=v \cdot \log v \Rightarrow \frac{d v}{v \log v}=\frac{d x}{x}$

Integrating both sides, we get

$ \qquad \int \frac{d v}{v \log v}=\int \frac{d x}{x} $

Put $\log v=t$ on L.H.S.

$ \begin{matrix} \frac{1}{v} d v =dt \\ \therefore \quad \int \frac{d t}{t} =\int \frac{d x}{x} \\ \log |t| =\log |x|+\log |C| \\ \Rightarrow \quad \log |\log v| =\log |x C| \Rightarrow |\log v|=|x C| \\ \Rightarrow \quad \log (\frac{y}{x}) =\pm Cx\\ \Rightarrow \quad \log (\frac{y}{x}) =cx, \quad where \quad c = \pm C \end{matrix} $

Hence, the required solution is $\log (\frac{y}{x})=x c$.

Solution

The degree of the given differential equation is not defined because the value of $\sin (\frac{d y}{d x})$ on expansion will be in increasing power of $(\frac{d y}{d x})$.

• Option (a) 1: This is incorrect because the degree of a differential equation is defined as the highest power of the highest order derivative, provided the equation is polynomial in derivatives. In this case, the equation involves a trigonometric function of the derivative, making it non-polynomial.

• Option (b) 2: This is incorrect because, although the highest power of the second-order derivative is 2, the presence of the trigonometric function $\sin (\frac{d y}{d x})$ means the equation is not polynomial in its derivatives, so the degree is not defined.

• Option (c) 3: This is incorrect because, although the highest power of the second-order derivative is 2, the presence of the trigonometric function $\sin (\frac{d y}{d x})$ means the equation is not polynomial in its derivatives, so the degree is not defined.

Solution

The given differential equation is

$ [1+(\frac{d y}{d x})^{2}]^{3 / 2}=(\frac{d^{2} y}{d x^{2}}) $

Squaring both sides, we have

$ [1+(\frac{d y}{d x})^{2}]^{3}=(\frac{d^{2} y}{d x^{2}})^{2} $

So, the degree of the given differential equation is 2 .

Hence, the correct option is $(d)$.

Solution

Given that, $\quad \frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^{1 / 4}=-x^{1 / 5}$

$ \begin{aligned} & \Rightarrow \quad \frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^{1 / 4}=-x^{1 / 5} \ & \Rightarrow \quad\left(\frac{d y}{d x}\right)^{1 / 4}=-\left(x^{1 / 5}+\frac{d^2 y}{d x^2}\right) \end{aligned} $

On squaring both sides, we get

$ \left(\frac{d y}{d x}\right)^{1 / 2}=\left(x^{1 / 5}+\frac{d^2 y}{d x^2}\right)^2 $

Again, on sqaring both sides, we have

$ \frac{d y}{d x}=\left(x^{1 / 5}+\frac{d^2 y}{d x^2}\right)^4 $

order $=2$, degree $=4$

Hence, the correct option is $(a)$.

Solution

Given equation is $y=e^{-x}(A \cos x+B \sin x)$

Differentiating both sides, w.r.t. $x$, we get

$ \begin{aligned} & \frac{d y}{d x}=e^{-x}(-A \sin x+B \cos x)-e^{-x}(A \cos x+B \sin x) \\ & \frac{d y}{d x}=e^{-x}(-A \sin x+B \cos x)-y \end{aligned} $

Again differentiating w.r.t. $x$, we get

$ \begin{aligned} & \frac{d^{2} y}{d x^{2}}=e^{-x}(-A \cos x-B \sin x)-e^{-x}(-A \sin x+B \cos x)-\frac{d y}{d x} \\ \Rightarrow & \frac{d^{2} y}{d x^{2}}=-e^{-x}(A \cos x+B \sin x)-[\frac{d y}{d x}+y]-\frac{d y}{d x} \\ \Rightarrow & \frac{d^{2} y}{d x^{2}}=-y-\frac{d y}{d x}-y-\frac{d y}{d x} \\ \Rightarrow & \frac{d^{2} y}{d x^{2}}=-2 \frac{d y}{d x}-2 y \Rightarrow \frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}+2 y=0 \end{aligned} $

Hence, the correct option is (c)

Solution

Given equation is : $y=A \cos \alpha x+B \sin \alpha x$ Differentiating both sides w.r.t. $x$, we have

$ \begin{aligned} \frac{d y}{d x} & =-A \sin \alpha x \cdot \alpha+B \cos \alpha x \cdot \alpha \\ & =-A \alpha \sin \alpha x+B \alpha \cos \alpha x \end{aligned} $

Again differentiating w.r.t. $x$, we get

$ \begin{aligned} \frac{d^{2} y}{d x^{2}} & =-A \alpha^{2} \cos \alpha x-B \alpha^{2} \sin \alpha x \\ \Rightarrow \quad \frac{d^{2} y}{d x^{2}} & =-\alpha^{2}(A \cos \alpha x+B \sin \alpha x) \\ \Rightarrow \quad \frac{d^{2} y}{d x^{2}} & =-\alpha^{2} y \Rightarrow \frac{d^{2} y}{d x^{2}}+\alpha^{2} y=0 \end{aligned} $

Hence, the correct option is (b).

Solution

The given differential equation is

$ x d y-y d x=0 $

$ \Rightarrow \quad \frac{d y}{d x}=\frac{y}{x} \Rightarrow \frac{d y}{y}=\frac{d x}{x} $

Integrating both sides, we get

$ \begin{aligned} & \qquad \int \frac{d y}{y}=\int \frac{d x}{x} \\ & \Rightarrow \quad \log |y|=\log |x|+\log|c| \Rightarrow \log |y|=\log |x c| \\ & \Rightarrow \quad |y|=|x c| \\ & \Rightarrow \quad y =\pm cx \\ & \Rightarrow \quad y =Cx \quad where \quad C = \pm c
& \text{ which is a straight line passing through the origin. } \\ & \text{ Hence, the correct answer is }(c) \end{aligned} $

Solution

The given differential equation is

$ \begin{aligned} \cos x \cdot \frac{d y}{d x}+y \sin x =1 \\ \Rightarrow \qquad \frac{d y}{d x}+\frac{\sin x}{\cos x} y =\frac{1}{\cos x} \Rightarrow \frac{d y}{d x}+\tan x y=\sec x \end{aligned} $

Here, $P=\tan x$ and $Q=\sec x$

$\therefore$ Integrating factor $=e^{\int P d x}=e^{\int \tan x d x}=e^{\log |\sec x|}=\sec x ,\quad (\sec x >0)$

Hence, the correct option is (c).

Solution

The given differential equation is

$ \tan y \sec ^{2} x d x+\tan x \sec ^{2} y d y=0 $

$\Rightarrow \tan x \sec ^{2} y d y=-\tan y \sec ^{2} x d x$

$\Rightarrow \quad \frac{\sec ^{2} y}{\tan y} \cdot d y=\frac{-\sec ^{2} x}{\tan x} \cdot d x$

Integrating both sides, we get

$ \begin{aligned} & \Rightarrow \quad \int \frac{\sec ^{2} y}{\tan y} d y=\int \frac{-\sec ^{2} x}{\tan x} d x \\ & \Rightarrow \quad \log |\tan y|=-\log |\tan x|+\log |c| \\ & \Rightarrow \quad \log |\tan y|+\log |\tan x|=\log |c| \\ & \Rightarrow \quad \log |\tan x. \tan y|=\log |c| \\ & \Rightarrow \quad |\tan x. \tan y|= |c| \\ & \Rightarrow \quad \tan x. \tan y= \pm c \\ & \Rightarrow \quad \tan x. \tan y= k \quad (where \quad k = \pm c) \end{aligned} $

Hence, option (d) is correct.

Solution

The given equation is

$$ \begin{equation*} y=A x+A^{3}\tag{1} \end{equation*} $$

Differentiating both sides, we get $\frac{d y}{d x}=A$

put the value of A in equation 1

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

So the degree of the differential equation is 3 .

Hence, the correct answer is (c).

Solution

The given differential equation is

$ x \frac{d y}{d x}-y=x^{4}-3 x \Rightarrow \frac{d y}{d x}-\frac{y}{x}=x^{3}-3 $

Here, $P=-\frac{1}{x}$ and $Q=x^{3}-3$

So, integrating factor $=e^{\int P d x}=e^{\int-\frac{1}{x} d x}=e^{-\log| x|}=e^{\log \frac{1}{x}}=\frac{1}{x} , (x>0) $

Hence, the correct option is (c).

Solution

The given differential equation is

$ \frac{d y}{d x}-y=1 $

Here, $P=-1, Q=1$

$\therefore$ Integrating factor, I.F. $=e^{\int P d x}=e^{\int-1 d x}=e^{-x}$

So, the solution is

$ \begin{aligned} & y \times \text{ I.F. }=\int Q \times \text{ I.F. } d x+c \\ & \Rightarrow \quad y \times e^{-x}=\int 1 . e^{-x} d x+c \\ & \Rightarrow \quad y \cdot e^{-x}=-e^{-x}+c \\ & \text{ Put } x=0, y=1 \\ & \Rightarrow \quad 1 \cdot e^{0}=-e^{0}+c \\ & \Rightarrow \quad 1=-1+c \quad \\ & \therefore c=2 \end{aligned} $

So the equation is $y \cdot e^{-x}=-e^{-x}+2$

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

Hence, the correct option is $(d)$.

Solution

The given differential equation is $\frac{d y}{d x}=\frac{y+1}{x-1}$

$ \Rightarrow \quad \frac{d y}{y+1}=\frac{d x}{x-1} $

Integrating both sides, we get

$\int \frac{d y}{y+1}=\int \frac{d x}{x-1}$

$\Rightarrow \log |(y+1)|=\log |(x-1)|+\log |c| $

$\Rightarrow \quad \log |(y+1)|-\log |(x-1)|=\log |c|$

$\Rightarrow \log |\frac{y+1}{x-1}|=\log |c| $

$\Rightarrow \frac{y+1}{x-1}=C \text{ Put } x=1 \text{ and } y=2 \quad where \quad C = \pm c$

$\Rightarrow \frac{2+1}{1-1}=C \quad \therefore C=\infty$

$\therefore \frac{y+1}{x-1}=\frac{1}{0} \Rightarrow x-1=0 \Rightarrow x=1$

Hence, the correct option is $(b)$.

Solution

Second order differential equation is $y^{\prime} y^{\prime \prime}+y=\sin x$

Hence, the correct option is $(b)$.

Solution

The given differential equation is

$ \begin{aligned} & (1-x^{2}) \frac{d y}{d x}-x y=1 \\ \Rightarrow \quad & \frac{d y}{d x}-\frac{x}{1-x^{2}} \cdot y=\frac{1}{1-x^{2}} \end{aligned} $

Here, $P=-\frac{x}{1-x^{2}}$ and $Q=\frac{1}{1-x^{2}}$

$\therefore$ Integrating factor

$ \text{ I.F. }=e^{\int P d x}=e^{\int \frac{-x}{1-x^{2}} d x}=e^{\frac{1}{2} \log |(1-x^{2})|}=\sqrt{1-x^{2}}, \quad ( (1-x^{2})>0) $

Hence, the correct option is (c).

Solution

Given equation is $\tan ^{-1} x+\tan ^{-1} y=c$

Differentiating w.r.t. $x$, we have

$ \begin{aligned} \frac{1}{1+x^{2}}+\frac{1}{1+y^{2}} \cdot \frac{d y}{d x} & =0 \\ \Rightarrow \quad(\frac{1}{1+y^{2}}) \frac{d y}{d x}=-(\frac{1}{1+x^{2}}) & \Rightarrow \frac{d y}{d x}=-(\frac{1+y^{2}}{1+x^{2}}) \\ \Rightarrow \quad(1+x^{2}) d y & =-(1+y^{2}) d x \\ \Rightarrow \quad(1+x^{2}) d y+(1+y^{2}) d x & =0 \end{aligned} $

Hence the correct option is (c).

Solution

Given differential equation is

$ \begin{aligned} y \frac{d y}{d x}+x & =c \\ \Rightarrow \quad y \frac{d y}{d x} & =c-x \quad \Rightarrow y d y=(c-x) d x \end{aligned} $

$\therefore$ Integrating both sides, we get

$\int y d y =\int(c-x) d x$

$\Rightarrow \frac{y^{2}}{2} =c x-\frac{x^{2}}{2}+k\Rightarrow \frac{x^{2}}{2}+\frac{y^{2}}{2}-c x=k$

$\Rightarrow x^{2}+y^{2}-2 c x =2 k \text{ which is a family of circles. }$

Hence, the correct option is (d).

Solution

The given differential equation is

$ e^{x} \cos y d x-e^{x} \sin y d y=0 $

$\Rightarrow e^{x}(\cos y d x-\sin y d y)=0$

$\Rightarrow \quad \cos y d x-\sin y d y=0$

$\Rightarrow \quad \sin y d y=\cos y d x \Rightarrow \frac{\sin y}{\cos y} d y=d x$

Integrating both sides, we have

$ \begin{aligned} \int \frac{\sin y}{\cos y} d y & =\int d x \\ \Rightarrow \quad-\log |\cos y| & =x+\log k \Rightarrow \log |\frac{1}{\cos y}|-\log k=x \\ \Rightarrow \quad \log (\frac{1}{k |\cos y|}) & =x \Rightarrow \frac{1}{k |\cos y|}=e^{x} \\ \Rightarrow \quad \frac{1}{k} & =e^{x} |\cos y| \Rightarrow e^{x} \cos y= \quad \pm \frac{1}{k} \\ &\Rightarrow e^{x} \cos y= c, \quad (where \quad c = \pm \frac {1}{k} ) \end{aligned} $

Hence, the correct option is (a).

Solution

The degree of the given differential equation is 1 as the power of the highest order is 1.

Hence, the correct option is $(a)$.

Solution

The given differential equation is

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

Since, it is a linear differential equation of first order.

$\therefore P=1$ and $Q=e^{-x}$

$\therefore$ I.F $=e^{\int 1 . d x}=e^{x}$

So, the solution is

$ \begin{aligned} & \quad y \times \text{ I.F. }= \int Q \times \cdot \text{ I.F. } d x+c \Rightarrow y \cdot e^{x}=\int e^{-x} \cdot e^{x} d x+c \\ & \Rightarrow \quad y \cdot e^{x}=\int 1 \cdot d x+c \Rightarrow y \cdot e^{x}=x+c \\ & \text{ Put } x=0, y=0, \text{ we have } 0=0+c \quad \\ & \therefore c=0 \end{aligned} $

So, the solution is $y e^{x}=x \Rightarrow y=x \cdot e^{-x}$

Hence, the correct option is (b).

Solution

Given differential equation is

$ \frac{d y}{d x}+y \tan x-\sec x=0 \Rightarrow \frac{d y}{d x}+y \tan x=\sec x $

Here, $P=\tan x$ and $Q=\sec x$

$\therefore$ I.F. $=e^{\int P d x}=e^{\int \tan x d x}=e^{\log |\sec x|}=\sec x , (\sec x>0)$

Hence, the correct option is (b).

Solution

The given differential equation is

$ \frac{d y}{d x}=\frac{1+y^{2}}{1+x^{2}} \Rightarrow \frac{d y}{1+y^{2}}=\frac{d x}{1+x^{2}} $

Integrating both sides, we get

$\int \frac{d y}{1+y^{2}} =\int \frac{d x}{1+x^{2}}$

$\Rightarrow \tan ^{-1} y =\tan ^{-1} x+c \Rightarrow \tan ^{-1} y-\tan ^{-1} x=c$

$\Rightarrow \tan^{-1}(\frac{y-x}{1+x y}) =c$

$\Rightarrow \frac{y-x}{1+x y} =\tan c \Rightarrow \frac{y-x}{1+x y}=k \quad[k=\tan c]$

$\Rightarrow y-x =k(1+x y)$

Hence, the correct option is $(b)$.

Solution

The given differential equation is

$ \frac{d y}{d x}+y=\frac{1+y}{x} $

$\begin{matrix} \Rightarrow & \frac{d y}{d x}=\frac{1+y}{x}-y \\ \Rightarrow & \frac{d y}{d x}=\frac{1}{x}+y \frac{(1-x)}{x} \Rightarrow \frac{d y}{d x}-(\frac{1-x}{x}) y=\frac{1}{x}\end{matrix} $

Here, $P=-(\frac{1-x}{x})$ and $Q=\frac{1}{x}$

$\therefore$ Integrating factor I.F. $=e^{\int P d x}=e^{\int \frac{x-1}{x} d x}=e^{\int(1-\frac{1}{x}) d x}$

$ =e^{(x-\log |x|)}=e^{x} \cdot e^{-\log x}, (x>0) $

$ =e^{x} \cdot e^{\log \frac{1}{x}}=e^{x} \cdot \frac{1}{x} $

Hence, the correct option is $(b)$.

Solution

The given equation is $y=a e^{m x}+b e^{-m x}$

On differentiation, we get $\frac{d y}{d x}=a \cdot m e^{m x}-b \cdot m e^{-m x}$

Again differentiating w.r.t., we have

$ \begin{aligned} & \frac{d^{2} y}{d x^{2}}=a m^{2} e^{m x}+b m^{2} e^{-m x} \\ \Rightarrow & \frac{d^{2} y}{d x^{2}}=m^{2}(a e^{m x}+b e^{-m x}) \Rightarrow \frac{d^{2} y}{d x^{2}}=m^{2} y \Rightarrow \frac{d^{2} y}{d x^{2}}-m^{2} y=0 \end{aligned} $

Hence, the correct option is (c).

Solution

The given differential equation is $\cos x \sin y d x+\sin x \cos y d y=0$

$\Rightarrow \quad \sin x \cos y d y=-\cos x \sin y d x$

$\Rightarrow \frac{\cos y}{\sin y} d y=-\frac{\cos x}{\sin x} d x \Rightarrow \cot y d y=-\cot x d x$

Integrating both sides, we have

$\Rightarrow \quad \int \cot y d y=-\int \cot x d x$

$ \begin{aligned} & \Rightarrow \quad \log |\sin y|=-\log |\sin x|+\log |C| \\ & \Rightarrow \quad \log |\sin y|+\log |\sin x|=\log |C| \\ & \Rightarrow \quad \log |\sin y \cdot \sin x|=\log |C| \Rightarrow \sin x \sin y=\pm C \\ & \Rightarrow \sin x \sin y=c, \quad (where \quad c= \pm C) \end{aligned} $

Hence, the correct option is (b).

Solution

The given differential equation is $x \frac{d y}{d x}+y=e^{x}$

$\Rightarrow \quad \frac{d y}{d x}+\frac{y}{x}=\frac{e^{x}}{x}$

Here $P=\frac{1}{x}$ and $Q=\frac{e^{x}}{x}$

$\therefore$ Integrating factor I.F. $=e^{\int \frac{1}{x} d x}=e^{\log |x|}=x, (x>0)$

So, the solution is

$ \begin{aligned} y \times \text{ I.F. } & =\int Q \times \text{ I.F. } d x+k \Rightarrow y \times x=\int \frac{e^{x}}{x} \times x d x+k \\ \Rightarrow y \times x & =\int e^{x} d x+k \Rightarrow y \times x=e^{x}+k \\ \therefore y & =\frac{e^{x}}{x}+\frac{k}{x} \end{aligned} $

Hence, the correct option is $(a)$.

Solution

The given equation is

$$ \begin{equation*} x^{2}+y^{2}-2 a y=0 \tag{1} \end{equation*} $$

Differentiating w.r.t. $x$, we have

$ \begin{aligned} & 2 x+2 y \cdot \frac{d y}{d x}-2 a \frac{d y}{d x}=0 \\ & \Rightarrow \quad x+y \frac{d y}{d x}-a \frac{d y}{d x}=0 \quad \Rightarrow \quad x+(y-a) \frac{d y}{d x}=0 \\ & \Rightarrow \quad(y-a) \frac{d y}{d x}=-x \quad \Rightarrow y-a=\frac{-x}{d y / d x} \end{aligned} $

$\Rightarrow \quad a=y+\frac{x}{d y / d x} \Rightarrow a=\frac{y \cdot \frac{d y}{d x}+x}{\frac{d y}{d x}}$

Putting the value of $a$ in eq. (1) we get

$ \begin{aligned} & x^{2}+y^{2}-2 y[\frac{y \frac{d y}{d x}+x}{\frac{d y}{d x}}]=0 \\ & \Rightarrow \quad(x^{2}+y^{2}) \frac{d y}{d x}-2 y(y \frac{d y}{d x}+x)=0 \\ & \Rightarrow \quad(x^{2}+y^{2}) \frac{d y}{d x}-2 y^{2} \frac{d y}{d x}-2 x y=0 \\ & \Rightarrow \quad(x^{2}+y^{2}-2 y^{2}) \frac{d y}{d x}=2 x y \Rightarrow(x^{2}-y^{2}) \frac{d y}{d x}=2 x y \end{aligned} $

$\therefore$ Hence the correct option is $(a)$.

Solution

The given equation is

$$ \begin{equation*} y=A x+A^{3}\tag{1} \end{equation*} $$

Differentiating both sides, we get $\frac{d y}{d x}=A$

put the value of A in equation 1

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

So the order of the differential equation is 1 .

Hence, the correct option is $(c)$.

Solution

The given differential equation is

$ \begin{aligned} \frac{d y}{d x} & =2 x \cdot e^{x^{2}-y} \\ \Rightarrow \quad \frac{d y}{d x} & =2 x \cdot e^{x^{2}} \cdot e^{-y} \Rightarrow \frac{d y}{e^{-y}}=2 x \cdot e^{x^{2}} d x \end{aligned} $

Integrating both sides, we have

$ \begin{aligned} \int \frac{d y}{e^{-y}}=\int 2 x \cdot e^{x^{2}} d x \Rightarrow & \int e^{y} d y=\int 2 x \cdot e^{x^{2}} d x \\ & \text{ Put in RHS } x^{2}=t \therefore 2 x d x=d t \end{aligned} $

$ \begin{aligned} \therefore \quad \int e^{y} d y & =\int e^{t} d t \\ \Rightarrow \quad e^{y} & =e^{t}+c \Rightarrow e^{y}=e^{x^{2}}+c \\ \Rightarrow \quad e^{y-x^{2}} = c \end{aligned} $

Hence, the correct option is (c).

Solution

Since, the slope of the tangent to the curve $=x: y$

$ \therefore \quad \frac{d y}{d x}=\frac{x}{y} \Rightarrow y d y=x d x $

Integrating both sides, we get $\int y d y=\int x d x$

$ \begin{aligned} & \Rightarrow \quad \frac{y^{2}}{2}=\frac{x^{2}}{2}+c \Rightarrow y^{2}=x^{2}+2 c \\ & \Rightarrow \quad y^{2}-x^{2}=2 c=k \text{ which is rectangular hyperbola. } \end{aligned} $

Hence, the correct option is $(d)$.

Solution

The given differential equation is

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

Since it is linear differential equation where $P=-x$ and $Q=e^{\frac{x^{2}}{2}}$

$\therefore$ Integrating factor I.F. $=e^{\int P d x}=e^{\int-x d x}=e^{-\frac{x^{2}}{2}}$

So, the solution is

$ \begin{aligned} & y \times \text{ I.F. }=\int Q \times \text{ I.F. } d x+c \\ & \Rightarrow \quad y \times e^{-\frac{x^{2}}{2}}=\int e^{\frac{x^{2}}{2}} e^{-\frac{x^{2}}{2}} d x+c \\ & \Rightarrow \quad y \times e^{-\frac{x^{2}}{2}}=\int e^{0} d x+c \\ & \Rightarrow \quad y \times e^{-\frac{x^{2}}{2}}=\int 1 . d x+c \Rightarrow y \times e^{-\frac{x^{2}}{2}}=x+c \end{aligned} $

$ \therefore \quad y=(x+c) e^{\frac{x^{2}}{2}} $

Hence the correct option is $(c)$.

Take left hand side

$\frac{d y}{d x} = \frac{d ((c-x) e^{\frac{x^{2}}{2}})}{d x}$

$=(c-x)\cdot e^\frac {x^{2}}{2} \cdot \frac {2x}{2}+e^\frac {x^{2}}{2} (0-1) $

$=x((c-x)e^\frac {x^{2}}{2})-e^\frac {x^{2}}{2}$

$=xy-e^\frac {x^{2}}{2}$

$\therefore $ the left-hand side is not equal the right-hand side.

Hence the option (d) is incorrect.

Solution

The given differential equation is

$ \begin{aligned} (2 y-1) d x-(2 x+3) d y & =0 \Rightarrow(2 x+3) d y=(2 y-1) d x \\ \Rightarrow \quad \frac{d y}{2 y-1} & =\frac{d x}{2 x+3} \end{aligned} $

Integrating both sides, we get

$ \begin{matrix} \int \frac{d y}{2 y-1} =\int \frac{d x}{2 x+3} \\ \Rightarrow \quad \frac{1}{2} \log |2 y-1| =\frac{1}{2} \log |2 x+3|+\log c \\ \Rightarrow \quad \log |2 y-1| =\log |2 x+3|+2 \log c \\ \Rightarrow \quad \log |2 y-1|-\log |2 x+3| =\log c^{2} \\ \Rightarrow \quad \log |\frac{2 y-1}{2 x+3}| =\log c^{2} \\ \Rightarrow \quad \frac{2 y-1}{2 x+3} = \pm c^{2}\\ \Rightarrow \quad \frac{2 y-1}{2 x+3} = C , \quad where \quad C = \pm c^2 \\ \Rightarrow \quad \frac{2 x+3}{2 y-1}=\frac{1}{C} \\ \Rightarrow \quad \frac{2 x+3}{2 y-1} =k \text{, where } k=\frac{1}{C} \end{matrix} $

Hence, the correct option is (c).

Solution

The given equation is

$ \begin{gathered} y=a \cos x+b \sin x \\ \frac{d y}{d x}=-a \sin x+b \cos x \end{gathered} $

$ \begin{aligned} \frac{d^{2} y}{d x^{2}} & =-a \cos x-b \sin x \\ \Rightarrow \quad \frac{d^{2} y}{d x^{2}} & =-(a \cos x+b \sin x) \Rightarrow \frac{d^{2} y}{d x^{2}}=-y \Rightarrow \frac{d^{2} y}{d x^{2}}+y=0 \end{aligned} $

Hence, the correct option is $(a)$.

Solution

The given differential equation is $\frac{d y}{d x}+y=e^{-x}$

Since, it is a linear differential equation then $P=1$ and $Q=e^{-x}$

Integrating factor I.F. $=e^{\int P d x}=e^{\int 1 . d x}=e^{x}$

$\therefore$ Solution is

$ \begin{aligned} y \times \text{ I.F. } & =\int Q \times \text{ I.F. } d x+c \\ \Rightarrow \quad y \times e^{x} & =\int e^{-x} \times e^{x} d x+c \Rightarrow y \times e^{x}=\int e^{0} d x+c \\ \Rightarrow \quad y \times e^{x} & =\int 1 . d x+c \Rightarrow y \times e^{x}=x+c \end{aligned} $

Put $y=0$ and $x=0$

$\therefore \quad 0=0+c \quad $

$\therefore \quad c=0$

$\therefore$ equation is $y \times e^{x}=x$

So $\quad y=x \cdot e^{-x}$

Hence, the correct option is $(d)$.

Solution

The given differential equation is

$ [\frac{d^{3} y}{d x^{3}}]^{2}-3 \frac{d^{2} y}{d x^{2}}+2(\frac{d y}{d x})^{4}=y^{4} $

Here the highest derivative is $\frac{d^{3} y}{d x^{3}}$.

$\therefore$ the order of the differential equation is 3 and since, the power of highest order is 2

$\therefore$ its degree is 2

Hence, the correct option is (d).

Solution

The given differential equation is

$ [1+(\frac{d y}{d x})^{2}]=\frac{d^{2} y}{d x^{2}} $

Here, the highest derivative is 2 ,

$\therefore$ order $=2$

and the power of the highest derivative is 1

$\therefore$ degree $=1$

Hence, the correct option is (c).

Solution

The given equation of family of curves is

$$ \begin{align*} & y^{2}=4 a(x+a) \\ \Rightarrow \quad & y^{2}=4 a x+4 a^{2} \tag{1} \end{align*} $$

Differentiating both sides, w.r.t. $x$, we get

$ \begin{aligned} 2 y \cdot \frac{d y}{d x} & =4 a \\ \Rightarrow \quad y \cdot \frac{d y}{d x} & =2 a \Rightarrow \frac{y}{2} \frac{d y}{d x}=a \end{aligned} $

Now, putting the value of $a$ in eq. (1) we get

$ \begin{aligned} y^{2} & =4 x(\frac{y}{2} \frac{d y}{d x})+4(\frac{y}{2} \cdot \frac{d y}{d x})^{2} \\ \Rightarrow \quad y^{2} & =2 x y \frac{d y}{d x}+y^{2}(\frac{d y}{d x})^{2} \Rightarrow y=2 x \frac{d y}{d x}+y(\frac{d y}{d x})^{2} \end{aligned} $

$ \Rightarrow \quad 2 x \cdot \frac{d y}{d x}+y \cdot(\frac{d y}{d x})^{2}-y=0 $

Hence, the correct option is (d).

Solution

The given differential equation is

$ \frac{d^{2} y}{d x^{2}}-2 \frac{d y}{d x}+y=0 $

Since the above equation is of second order and first degree

$\therefore \quad D^{2} y-2 D y+y=0$, where $D=\frac{d}{d x}$

$\Rightarrow(D^{2}-2 D+1) y=0$

$\therefore$ auxiliary equation is

$ m^{2}-2 m+1=0 \Rightarrow(m-1)^{2}=0 \Rightarrow m=1,1 $

If the roots of Auxiliary equation are real and equal say $(m)$ then $C F=(ax+b) \cdot e^{m x}$

$\therefore \quad C F=(A x+B) e^{x}$

So $y=(A x+B) \cdot e^{x}$

Hence, the correct option is $(a)$.

Solution

The given differential equation is $\frac{d y}{d x}+y \tan x=\sec x$

Since, it is a linear differential equation

$\therefore \quad P=\tan x$ and $Q=\sec x$

Integrating factor I.F. $=e^{\int P d x}=e^{\int \tan x d x}=e^{\log |\sec x|}=\sec x, (\sec x>0)$

$\therefore$ Solution is

$ \begin{aligned} y \times \text{ I.F. } & =\int Q \times \text{ I.F. } d x+c \\ \Rightarrow \quad y \times \sec x & =\int \sec x \cdot \sec x d x+c \\ \Rightarrow \quad y \sec x & =\int \sec ^{2} x d x+c \Rightarrow y \sec x=\tan x+c \\ \Rightarrow y \sec x=\tan x+c \end{aligned} $

Hence, the correct option is (a).

Solution

The given differential equation is $\frac{d y}{d x}+\frac{y}{x}=\sin x$

Since, it is a linear differential equation

$\therefore \quad P=\frac{1}{x}$ and $Q=\sin x$

Integrating factor I.F. $=e^{\int \frac{1}{x} d x}=e^{\log |x|}=x , (x>0)$

$\therefore$ Solution is $y \times$ I.F. $=\int Q \times$ I.F. $d x+c$

$ \begin{matrix} y \times x =\int \sin x \cdot x d x+c \Rightarrow y \times x=\int x \sin x d x+c \\ y x =x \cdot \int \sin x d x-\int(D(x) \int \sin x d x) d x+c \\ \Rightarrow y x =x(-\cos x)-\int-\cos x d x \\ \Rightarrow y x=-x \cos x+\int \cos x d x \Rightarrow y x=-x \cos x+\sin x+c \\ \Rightarrow y x+x \cos x =\sin x+c \\ \Rightarrow x(y+\cos x) =\sin x+c \end{matrix} $

Hence, the correct option is $(a)$.

Solution

The given differential equation is

$ \begin{aligned} (e^{x}+1) y d y & =(y+1) e^{x} d x \\ \Rightarrow \quad \frac{y}{y+1} d y & =\frac{e^{x}}{e^{x}+1} d x \end{aligned} $

Integrating both sides, we get

$ \begin{aligned} \int \frac{y}{y+1} d y & =\int \frac{e^{x}}{e^{x}+1} d x \\ \Rightarrow \quad \int \frac{y+1-1}{y+1} d y & =\int \frac{e^{x}}{e^{x}+1} d x \\ \Rightarrow \quad \int 1 \cdot d y-\int \frac{1}{y+1} d y & =\int \frac{e^{x}}{e^{x}+1} d x \end{aligned} $

$ \begin{aligned} \Rightarrow \quad y-\log |y+1| =\log |e^{x}+1|+\log |c| \\ \Rightarrow \quad y =\log |y+1|+\log |e^{x}+1|+\log |c| \\ \Rightarrow \quad y =\log |c(y+1)(e^{x}+1)| \\ \Rightarrow \quad y =\log [k(y+1)(e^{x}+1)] , \quad (where \quad k= \pm c) \end{aligned} $

Hence, the correct option is (c).

Solution

The given differential equation is

$ \begin{aligned} & \frac{d y}{d x}=e^{x-y}+x^{2} e^{-y} \\ & \Rightarrow \quad \frac{d y}{d x}=e^{x} \cdot e^{-y}+x^{2} \cdot e^{-y} \Rightarrow \frac{d y}{d x}=e^{-y}(e^{x}+x^{2}) \\ & \Rightarrow \quad \frac{d y}{e^{-y}}=(e^{x}+x^{2}) d x \Rightarrow e^{y} \cdot d y=(e^{x}+x^{2}) d x \end{aligned} $

Integrating both sides, we have

$ \begin{aligned} \int e^{y} d y & =\int(e^{x}+x^{2}) d x \\ \Rightarrow \quad e^{y} & =e^{x}+\frac{x^{3}}{3}+c \Rightarrow e^{y}-e^{x}=\frac{x^{3}}{3}+c \end{aligned} $

Hence, the correct option is $(b)$.

Solution

The given differential equation is

$ \frac{d y}{d x}+\frac{2 x y}{1+x^{2}}=\frac{1}{(1+x^{2})^{2}} $

Since, it is a linear differential equation

$ P=\frac{2 x}{1+x^{2}} \text{ and } Q=\frac{1}{(1+x^{2})^{2}} $

Integrating factor I.F. $=e^{\int P d x}=e^{\int \frac{2 x}{1+x^{2}} d x}=e^{\log |(1+x^{2})|}=|(1+x^{2})|=(1+x^{2}), \quad (\because (1+x^{2}) \quad is \quad positive) $

$\therefore$ Solution is $y \times$ I.F. $=\int Q \times$ I.F. $d x+c$

$ \begin{aligned} & \Rightarrow \quad y(1+x^{2})=\int \frac{1}{(1+x^{2})^{2}} \times(1+x^{2}) d x+c \\ & \Rightarrow \quad y(1+x^{2})=\int \frac{1}{(1+x^{2})} d x+c \Rightarrow y(1+x^{2})=\tan ^{-1} x+c \end{aligned} $

Hence, the correct option is $(a)$.

Solution

(i) The degree of the differential equation $\frac{d^{2} y}{d x^{2}}+e^{d y / d x}=0$ is not defined.

(ii) The given differential equation is $\sqrt{1+(\frac{d y}{d x})^{2}}=x$ Squaring both sides, we get

$ 1+(\frac{d y}{d x})^{2}=x^{2} $

So, the degree of the equation is 2 .

(iii) The number of arbitrary constants in the solution is 3 .

(iv) The given differential equation $\frac{d y}{d x}+\frac{y}{x \log x}=\frac{1}{x}$ is of the type $\frac{d y}{d x}+P y=Q$.

(v) General solution of the differential equation of the type $\frac{d x}{d y}+P_1 x=Q_1$ is given by $x \times$ I.F. $=\int Q \times$ I.F. $d y+c$

$\Rightarrow \quad x \cdot e^{\int P_1 d y}=\int Q_1 \cdot e^{\int P_1 d y} d y+c$.

(vi) The given differential equation is $x \frac{d y}{d x}+2 y=x^{2}$

$\Rightarrow \quad \frac{d y}{d x}+\frac{2}{x} y=x$.

Since, it is linear differential equation

$\therefore \quad P=\frac{2}{x}$ and $Q=x$

Integrating factor I.F. $=e^{\int P d x}=e^{\int \frac{2}{x} d x}=e^{2 \log |x|}=e^{\log |x^{2}|}=x^{2}$

$\therefore$ Solution is

$ \begin{aligned} y \times \text{ I.F. } & =\int Q \times \text{ I.F. } d x+c \\ \Rightarrow \quad y \cdot x^{2} & =\int x \cdot x^{2} d x+c \quad \Rightarrow y \cdot x^{2}=\int x^{3} d x+c \\ \Rightarrow \quad y \cdot x^{2} & =\frac{1}{4} x^{4}+c \quad \Rightarrow y=\frac{1}{4} x^{2}+c \cdot x^{-2} \end{aligned} $

Hence, the solution is $y=\frac{1}{4} x^{2}+c \cdot x^{-2}$.

(vii) The given differential equation is

$ \begin{aligned} & (1+x^{2}) \frac{d y}{d x}+2 x y-4 x^{2}=0 \\ \Rightarrow \quad & \frac{d y}{d x}+\frac{2 x y}{1+x^{2}}=\frac{4 x^{2}}{1+x^{2}} \end{aligned} $

Since it is a linear differential equation

$\therefore \quad P=\frac{2 x}{1+x^{2}}$ and $Q=\frac{4 x^{2}}{1+x^{2}}$

Integrating factor I.F. $=e^{\int P d x}=e^{\int \frac{2 x}{1+x^{2}} d x}=e^{\log |(1+x^{2})|}=(1+x^{2}) , \quad (\because (1+x^{2}) \quad is \quad positive )$

$\therefore$ Solution is $y \times I$.F $=\int Q \times$ I.F. $d x+c$

$ \begin{aligned} \Rightarrow y \times(1+x^{2}) =\int \frac{4 x}{1+x^{2}} \times(1+x^{2}) d x+c \\ \Rightarrow y \times(1+x^{2}) =\int 4 x^{2} d x+c \Rightarrow y \times(1+x^{2})=\frac{4}{3} x^{3}+c \\ \Rightarrow y =\frac{4}{3} \frac{x^{3}}{(1+x^{2})}+c(1+x^{2})^{-1} \end{aligned} $

$\Rightarrow 3y (1+x^{2})= 4x^{3}+3c$

$\Rightarrow 3y (1+x^{2})= 4x^{3}+C , (where \quad C=3c)$

Hence, the required solution is $ 3y (1+x^{2})= 4x^{3}+C$

(viii) The given differential equation is

$ \begin{gathered} y d x+(x+x y) d y=0 \\ \Rightarrow \quad(x+x y) d y=-y d x \Rightarrow x(1+y) d y=-y d x \\ \Rightarrow \quad \frac{1+y}{y} d y=-\frac{1}{x} d x \end{gathered} $

Integrating both sides, we get

$ \begin{aligned} & \int \frac{1+y}{y} d y=-\int \frac{1}{x} d x \\ & \Rightarrow \quad \int(\frac{1}{y}+1) d y=-\int \frac{1}{x} d x \\ & \Rightarrow \quad \log |y|+y=-\log |x|+\log |c| \\ & \Rightarrow \log |x|+\log |y|=\log |c|-y \\ & \Rightarrow \quad \log |x y |=\log |c|-y \\ & \Rightarrow \quad |x y |=e^{\log|c|}e^{-y} \\ & \Rightarrow \quad x y =\pm e^{\log|c|}e^{-y} \\ & \Rightarrow \quad x y =Ce^{-y} , \quad (where \quad C=\pm e^{\log|c|} )\\ \end{aligned} $

Hence, the required solution is $x y=C e^{-y}$.

(ix) The given differential equation is $\frac{d y}{d x}+y=\sin x$

Since, it it a linear differential equation

$\therefore \quad P=1$ and $Q=\sin x$

Integrating factor I.F. $=e^{\int P d x}=e^{\int 1 . d x}=e^{x}$

$\therefore$ Solution is $y \times$ I.F. $=\int Q \times$ I.F. $d x+c$

$\Rightarrow \quad y \cdot e^{x}=\int \sin x \cdot e^{x} d x+c$

Let $ I=\int \sin x \cdot e^{x} d x$

$ Intigration \quad by \quad part $

$ \begin{aligned} & I=\sin x \cdot \int e^{x} d x-\int(D(\sin x) \cdot \int e^{x} d x) d x \\ & I=\sin x \cdot e^{x}-\int \cos x \cdot e^{x} d x \\ & I=\sin x \cdot e^{x}-[\cos x \cdot \int e^{x} d x-\int(D(\cos x) \int e^{x} d x) d x] \end{aligned} $

$ \begin{aligned} I & =\sin x \cdot e^{x}-[\cos x \cdot e^{x}-\int-\sin x \cdot e^{x} d x] \\ I & =\sin x \cdot e^{x}-\cos x \cdot e^{x}-\int \sin x \cdot e^{x} d x \\ I & =\sin x \cdot e^{x}-\cos x \cdot e^{x}-I \\ \Rightarrow I+I & =e^{x}(\sin x-\cos x) \\ \Rightarrow \quad 2 I & =e^{x}(\sin x-\cos x) \\ \therefore \quad I & =\frac{e^{x}}{2}(\sin x-\cos x) \end{aligned} $

From eq. (1) we get

$ \begin{aligned} y \cdot e^{x} & =\frac{e^{x}}{2}(\sin x-\cos x)+c \\ y & =(\frac{\sin x-\cos x}{2})+c \cdot e^{-x} \end{aligned} $

Hence, the required solution is

$ y=(\frac{\sin x-\cos x}{2})+c \cdot e^{-x} $

(x) The given differential equation is $\cot y d x=x d y$

$ \Rightarrow \frac{d y}{\cot y}=\frac{d x}{x} \Rightarrow \tan y d y=\frac{d x}{x} $

Integrating both sides, we get

$ \begin{aligned} & \quad \int \tan y d y=\int \frac{d x}{x} \Rightarrow \log |\sec y|=\log |x|+\log |c| \\ \Rightarrow \quad & \log |\sec y|-\log |x|=\log |c| \\ \Rightarrow \quad & \log |\frac{\sec y}{x}|=\log |c| \\ \therefore \quad & |\frac{\sec y}{x}|=|c| \\ \therefore \quad & \frac{\sec y}{x}=\pm c \\ \therefore \quad & \frac{\sec y}{x}=C \quad ( where \quad C = \pm c)\\ \Rightarrow & \frac{x}{\sec y}=\frac{1}{C} \Rightarrow \frac{x}{\sec y}=C \\ \therefore \quad & x=C \sec y \end{aligned} $

Hence, the required solution is $x=C \sec y$.

$(x i)$ The given differential equation is

$ \begin{aligned} \frac{d y}{d x}+y & =\frac{1+y}{x} \\ \frac{d y}{d x}+y & =\frac{1}{x}+\frac{y}{x} \\ \Rightarrow \quad \frac{d y}{d x}+y-\frac{y}{x} & =\frac{1}{x} \Rightarrow \frac{d y}{d x}+(1-\frac{1}{x}) y=\frac{1}{x} \end{aligned} $

$ \begin{aligned} & \text{ Here } P=(1-\frac{1}{x}) \\ & \therefore \quad \text{ I.F. }=e^{\int P d x}=e^{\int(1-\frac{1}{x}) d x}=e^{(x-\log |x|)} \\ & =e^{x} \cdot e^{-\log |x|}=e^{x} \cdot e^{\log |\frac{1}{x}|}=\frac{e^{x}}{|x|}=\frac{e^{x}}{x},(x>0) \end{aligned} $

Hence, the required I.F. $=\frac{e^{x}}{x}$

Solution

(i) True

I.F. of the given differential equation

$ \frac{d x}{d y}+P_1 x=Q \text{ is } e^{\int P_1 d y} $

(ii) True

(iii) True

(iv) True

(v) False

Since particular solution of a differential equation has no arbitrary constant.

(vi) False

We know that the order of the differential equation is equal to the number of arbitrary constants.

(vii) True

The given differential equation is

$ \begin{aligned} \frac{d y}{d x} & =(\frac{y}{x})^{1 / 3} \\ \Rightarrow \quad \frac{d y}{d x} & =\frac{y^{1 / 3}}{x^{1 / 3}} \Rightarrow \frac{d y}{y^{1 / 3}}=\frac{d x}{x^{1 / 3}} \end{aligned} $

Integrating both sides, we get

$\int \frac{d y}{y^{1 / 3}} =\int \frac{d x}{x^{1 / 3}} \Rightarrow \int y^{-1 / 3} d y=\int x^{-1 / 3} d x$

$\Rightarrow \frac{1}{-\frac{1}{3}+1} y^{-1 / 3+1} =\frac{1}{-1 / 3+1} \cdot x^{-1 / 3+1}+c$

$\Rightarrow \quad \frac{3}{2} y^{2 / 3} =\frac{3}{2} x^{2 / 3}+c$

$\Rightarrow \quad y^{2 / 3} =x^{2 / 3}+\frac{2}{3} c \Rightarrow y^{2 / 3}-x^{2 / 3}=k[k=\frac{2}{3} c]$

(viii) True

Given equation is

$ y=e^{x}(A \cos x+B \sin x) $

Differentiating both sides, we get

$ \begin{aligned} & \frac{d y}{d x}=e^{x}(-A \sin x+B \cos x)+(A \cos x+B \sin x) e^{x} \\ & \frac{d y}{d x}=e^{x}(-A \sin x+B \cos x)+y \end{aligned} $

Again differentiating w.r.t. $x$, we get

$ \begin{aligned} & \frac{d^{2} y}{d x^{2}}=e^{x}(-A \cos x-B \sin x)+(-A \sin x+B \cos x) \cdot e^{x}+\frac{d y}{d x} \\ & \frac{d^{2} y}{d x^{2}}=-e^{x}(A \cos x+B \sin x)+\frac{d y}{d x}-y+\frac{d y}{d x} \\ & \frac{d^{2} y}{d x^{2}}=-y-y+2 \frac{d y}{d x} \quad \therefore \quad \frac{d^{2} y}{d x^{2}}-2 \frac{d y}{d x}+2 y=0 \end{aligned} $

(ix) True

The given differential equation is

$ \begin{aligned} \frac{d y}{d x} & =\frac{x+2 y}{x} \\ \Rightarrow \quad \frac{d y}{d x} & =1+2 \frac{y}{x} \Rightarrow \frac{d y}{d x}-\frac{2 y}{x}=1 \end{aligned} $

Here, $P=\frac{-2}{x}$ and $Q=1$

Integrating factor I.F. $=e^{\int \frac{-2}{x} d x}=e^{-2 \log |x|}=e^{\log |x|^{-2}}=\frac{1}{|x|^{2}} = \frac{1}{x^{2}}$

$\therefore$ Solution is $y \times$ I.F. $=\int Q \times$ I.F. $d x+c$

$\Rightarrow y \times \frac{1}{x^{2}}=\int 1 \times \frac{1}{x^{2}} d x+c$

$\Rightarrow \quad \frac{y}{x^{2}}=\int \frac{1}{x^{2}} d x+c \Rightarrow \frac{y}{x^{2}}=-\frac{1}{x}+c$

$\Rightarrow \quad y=-x+c x^{2} \Rightarrow y+x=c x^{2}$

(x) True

The given differential equation is

$ \begin{aligned} & x \frac{d y}{d x}=y+x \tan (\frac{y}{x}) \\ & x \frac{d y}{d x}-x \tan (\frac{y}{x})=y \\ & \Rightarrow \frac{d y}{d x}-\tan (\frac{y}{x})=\frac{y}{x} \Rightarrow \frac{d y}{d x}=\frac{y}{x}+\tan (\frac{y}{x}) \\ & \text{ Put } y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x} \\ & \Rightarrow \quad v+x \cdot \frac{d v}{d x}=\frac{v x}{x}+\tan (\frac{v x}{x}) \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad v+x \frac{d v}{d x}=v+\tan v \Rightarrow x \frac{d v}{d x}=\tan v \\ & \Rightarrow \quad \frac{d v}{\tan v}=\frac{d x}{x} \Rightarrow \cot v d v=\frac{d x}{x} \end{aligned} $

Integrating both sides, we get

$ \begin{matrix} \int \cot v d v =\int \frac{d x}{x} \Rightarrow \log |\sin v|=\log |x|+\log |c| \\ \Rightarrow \log |\sin \frac{y}{x}|=\log |x c| \\ \therefore \qquad \sin \frac{y}{x} =\pm cx \\ \therefore \qquad \sin \frac{y}{x} =Cx , \quad (where \quad C= \pm c) \end{matrix} $

(xi) True

Let $y=m x+c$ be the non-horizontal line in a plane

$\therefore \qquad \frac{d y}{d x}=m$ and $\frac{d^{2} y}{d x^{2}}=0$.