Question

The solution of $\frac{dy}{dx}=\frac{y}{x}+sin\left(\frac{y}{x}\right)$ is:

• A. $\frac{1}{x}tan\left(\frac{y}{2x}\right)=c$
• B. $tan\left(\frac{y}{x}\right)-cx=0$
• C. $tan\left(\frac{y}{x}\right)+cx-x=0$
• D. $sin\left(\frac{y}{2x}\right)-cx=0$

Answer 1

The correct option is A $\frac{1}{x}tan\left(\frac{y}{2x}\right)=c$

Let $\frac{y}{x}=v$------(1)

$\Rightarrow y=xv$

$\Rightarrow \frac{dy}{dx}=vtx\frac{dv}{dx}$----------(2)

Substituting (1) and (2) in the given equation

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

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

$\Rightarrow cosecvdv=\frac{dx}{x}$

$\Rightarrow -log|cosecv+\cot v|+c=logx$

$-log|\frac{1}{\sin v}+\frac{\cos v}{\sin v}|c=logx$

$\Rightarrow -log\left|\frac{2{\cos }^{2}v/2}{2\sin v/2\cos /v/2}\right|=logcx$

$\Rightarrow \tan v/2-cx=0$ $\Rightarrow \tan \frac{y}{2x}-cx=0$