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 $c \log$ 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} \\ \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

Put $y=0$ and $x=5$

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

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

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

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

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

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

$\therefore$ Solution of the equation is

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

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}{d 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 \frac{y}{c}=e^{x-x^{2}} \\ & \therefore \quad y=y=c \cdot e^{x-x^{2}} \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 $=\int Q$ 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} $

Hence the required solution is $y=\frac{e^{m x}}{(m+a)}+c \cdot e^{-a x}$.

Solution

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

Put $x+y=t$

$ \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 $(x+c) \cdot e^{x} \quad y+1=0$.

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 c=\frac{x^{2}}{2} \Rightarrow \log \frac{x}{y c}=\frac{x^{2}}{2} \\ & \Rightarrow \quad \frac{x}{y c}=e^{x^{2} / 2} \Rightarrow \frac{y c}{x}=e^{-x^{2} / 2} \Rightarrow y c=x e^{-x^{2} / 2} \\ & \therefore \quad y=\frac{1}{c} \cdot x e^{-x^{2} / 2} \Rightarrow y=k x e^{-x^{2} / 2} \quad[\because k=\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 \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} $

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

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

So the solution of the equation is

$ \begin{aligned} x . I . F . & =\int \text{ Q.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 $x=y^{3}+c y=y(y^{2}+c)$

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 \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 \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} $

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

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

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

$ \begin{aligned} & =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)} \end{aligned} $

Required solution of the given differential equation is

$ \begin{matrix} y \cdot \text{ I.F. } =\int \text{ Q.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 \\ & \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

Now put $t=1$

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

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

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

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

$[\because y(0)=-1]$

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

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

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

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

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

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

$ 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} $

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

Integrating factor I.F. $=e^{\int P d x}=e^{\int \frac{2 x}{1+x^{2}} d x}=e^{\log (1+x^{2})}=1+x^{2}$

$\therefore$ 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}} $

Put $y=v x$

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

differential equation]

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

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

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

$ \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} $

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

Put $x=1, y=-2$

$\Rightarrow \quad \frac{1}{2}(-2)-\frac{3}{2} \log |-2+3|=\log (1)+c$

$\Rightarrow \quad-1-\frac{3}{2} \log (1)=\log (1)+c$

$\Rightarrow \quad-1-0=0+c \qquad[\because \log (1)=0]$

$\therefore \quad c=-1$

$\therefore$ equation is

$\quad \frac{1}{2} y-\frac{3}{2} \log |y+3| =\log x-1$

$\Rightarrow \quad y-3 \log |y+3| =2 \log x-2$

$\Rightarrow \quad y-\log |(y+3)^{3}| =\log x^{2}-2$

$\Rightarrow \quad \log |(y+3)^{3}|+\log x^{2} =y+2$

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

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

Solution

The given differential equation is

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

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

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

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

$\therefore$ Integrating factor I.F. $=e^{\int P d x}=e^{\int \cot x d x}=e^{\log \sin x}=\sin x$

$\therefore$ 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 $

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

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

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} $

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

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

$\therefore$ Solution is

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

Let I

$ \begin{aligned} & =\int(\sin x+\log x) x^{2} d x \\ & =\int x_I^{2} \sin _{II} x d x+\int _{II} x^{2} \log x d x \\ & \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} $

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) \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,

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

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

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

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

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

Integrating both sides, we have

$ \begin{aligned} \frac{d v}{\cos v+\sin v+1} =\int 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)} =\int 1 . d x \\ \Rightarrow \quad \int \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 =\int 1 . d x \\ \Rightarrow \quad \int \frac{\sec ^{2} \frac{v}{2}}{2+2 \tan ^{2}} d v =\int 1 . d x \\ \text{ Put } \quad 2+2 \tan \frac{v}{2} \end{aligned} $

$ \begin{matrix} 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 \end{matrix} $

Hence, the required solution is

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

Solution

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

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 . \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 _{I} 2 x \cdot e_II^{-3 x} d x \\ & \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} \iint_I^{\cos 2 x} \cdot e_II^{-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}-. \\ & .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}}=x 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} \Rightarrow(y-1)(x+1)=c x \end{matrix} $

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

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

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

$$ \begin{align*} \int \frac{d v}{1-v^{2}} & =\int d x \\ \Rightarrow \quad \frac{1}{2} \log |\frac{1+v}{1-v}| & =x+c \quad \Rightarrow \quad \frac{1}{2} \log |\frac{1+x-y}{1-x+y}|=x+c \tag{1} \end{align*} $$

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

then $\frac{1}{2} \log |\frac{1+0-0}{1-0+0}|=0+c \quad \Rightarrow \quad c=0$ $\therefore$ On putting $c=0$ in eq. (1) we get

$ \begin{aligned} & & \frac{1}{2} \log |\frac{1+x-y}{1-x+y}| & =x \Rightarrow \log |\frac{1+x-y}{1-x+y}| \\ & \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 \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

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

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

$ \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}) =x 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 $(\frac{d y}{d x})$.

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 differential equation is

$ \begin{aligned} & \frac{d^{2} y}{d x^{2}}+(\frac{d y}{d x})^{\frac{1}{4}}+x^{\frac{1}{5}}=0 \\ & \Rightarrow \quad \frac{d^{2} y}{d x^{2}}+(\frac{d y}{d x})^{\frac{1}{4}}=-x^{\frac{1}{5}} \end{aligned} $

Since the degree of $\frac{d y}{d x}$ is in fraction.

So, the degree of the differential equation is not defined as the order is 2.

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 \text{ which is a straight line passing } \\ & \text{ 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 $=e^{\int P d x}=e^{\int \tan x d x}=e^{\log \sec x}=\sec x$

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 \end{aligned} $

Solution

Given equation is $y=A x+A^{3}$

Differentiating both sides, we get

$ \frac{d y}{d x}=\text{ A which has degree } 1 \text{. } $

Hence, the correct answer is (a).

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 $=e^{\int P d x}=e^{\int-\frac{1}{x} d x}=e^{-\log x}=e^{\log \frac{1}{x}}=\frac{1}{x}$

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 \text{ Q. 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 $

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

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

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

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}} $

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=c \quad[c=\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

$\therefore P=1$ 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 \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. $=e^{\int P d x}=e^{\int \tan x d x}=e^{\log \sec x}=\sec x$

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} $

$ =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=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. $=e^{\int \frac{1}{x} d x}=e^{\log |x|}=x$

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

$ y=A x+A^{3} $

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

Again differentiating both sides, we have $\frac{d^{2} y}{d x^{2}}=0$

So the order of the differential equation is 2 .

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

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 \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 $(c)$.

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} =c^{2} \Rightarrow \frac{2 x+3}{2 y-1}=\frac{1}{c^{2}} \\ \Rightarrow \quad \frac{2 x+3}{2 y-1} =k \text{, where } k=\frac{1}{c^{2}} \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 y}{d x}+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 $C F=(c_1 x+c_2) \cdot e^{m x}$

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

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. $=e^{\int P d x}=e^{\int \tan x d x}=e^{\log \sec x}=\sec x$

$\therefore$ 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 \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. $=e^{\int \frac{1}{x} d x}=e^{\log x}=x$

$\therefore$ 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 & y-\log |y+1| & =\log |e^{x}+1|+\log k \\ \Rightarrow & y & =\log |y+1|+\log |e^{x}+1|+\log k \\ \Rightarrow & y & =\log |k(y+1)(e^{x}+1)| \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. $=e^{\int P d x}=e^{\int \frac{2 x}{1+x^{2}} d x}=e^{\log (1+x^{2})}=(1+x^{2})$

$\therefore$ 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 gett

$ 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 factorI.F. $=e^{\int P d x}=e^{\int \frac{2 x}{1+x^{2}} d x}=e^{\log (1+x^{2})}=(1+x^{2})$

$\therefore$ 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} $

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

(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 e^{y}=\log c \\ & \Rightarrow \quad \log (x y \cdot e^{y})=\log c \\ & \therefore \quad x y=c e^{-y} \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 $I=\int \sin _{I} x \cdot e^{x} d x$

$ \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 \Rightarrow \frac{x}{\sec y}=\frac{1}{C} \Rightarrow \frac{x}{\sec y}=C \quad[\frac{1}{c}=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}}=e^{x} \cdot \frac{1}{x} \end{aligned} $

Hence, the required I.F. $=e^{x} \cdot \frac{1}{x}$.

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. $=e^{\int \frac{-2}{x} d x}=e^{-2 \log x}=e^{\log x^{-2}}=\frac{1}{x^{2}}$

$\therefore$ 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 \quad \log \sin v-\log x =\log c \Rightarrow \log \sin \frac{y}{x}=\log x c \\ \therefore \qquad \sin \frac{y}{x} =x 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 $\frac{d^{2} y}{d x^{2}}=0$.