Question

The particular solution of the differential equation $\left[x{\sin }^2\left(\frac{y}{x}\right)-y\right]dx+xdy=0,$ satisfying the condition $y=\frac{\pi }{4},$ when $x=1$, is

• A. $\tan \left(\frac{y}{x}\right)=\log \left(ex\right)$
• B. $\tan \left(\frac{x}{y}\right)=\log \left(ex\right)$
• C. $\cot \left(\frac{y}{x}\right)=\log \left(\frac{e}{x}\right)$
• D. $\cot \left(\frac{y}{x}\right)=\log \left(ex\right)$

Answer 1

The correct option is D $\cot \left(\frac{y}{x}\right)=\log \left(ex\right)$

The given differential equation can be written as

$[x{\sin }^2\frac{y}{x}-y]dx+xdy=0$

$\Rightarrow xdy=-[x{\sin }^2\frac{y}{x}-y]dx$

$\Rightarrow \frac{dy}{dx}=-[{\sin }^2\frac{y}{x}-\frac{y}{x}]dx$

Let $F(x,y)=\frac{dy}{dx}=-[{\sin }^2\frac{y}{x}-\frac{y}{x}]dx$

$F(\lambda x,\lambda y)=-[{\sin }^2\frac{\lambda y}{\lambda x}-\frac{\lambda y}{\lambda x}]dx$

$\Rightarrow F(\lambda x,\lambda y)={\lambda }^{0}F(x,y)$

$\therefore F(x,y)$ is a homogeneous function of degree $0$

Put $y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$

$\Rightarrow v+x\frac{dv}{dx}=-[{\sin }^{2}v-v]$

$\Rightarrow v+x\frac{dv}{dx}=-{\sin }^{2}v+v$

$\Rightarrow x\frac{dv}{dx}=-{\sin }^{2}v$

$\Rightarrow -{\csc }^{2}vdv=\frac{dx}{x}$

Integrating both sides, we get

$\int -{\csc }^{2}vdv=\int \frac{dx}{x}$

$\Rightarrow \cot v=\log \left|x\right|+c$

$\Rightarrow \cot \frac{y}{x}-\log \left|x\right|=c$ where $v=\frac{y}{x}$

Put $x=1y=\frac{\pi }{4}$

$\Rightarrow \cot \frac{\frac{\pi }{4}}{1}-\log \left|1\right|=c$

$\Rightarrow \cot \frac{\pi }{4}-\log 1=c$

$\Rightarrow 1-0=c$

$\Rightarrow c=1$

$\Rightarrow \cot \frac{y}{x}-\log \left|x\right|=1$

$\Rightarrow \cot \frac{y}{x}-\log \left|x\right|=\log e$

$\Rightarrow \cot \frac{y}{x}=\log e+\log \left|x\right|$

$\Rightarrow \cot \frac{y}{x}=\log \left(ex\right)$