# Get Answers to all your Questions

### Answers (1)

Answer : $\tan \frac{y}{x}=-\log |x|+c$

Hint : You must know the rules of solving differential equation and integration

Given :     $x \frac{d y}{d x}+x \cos ^{2}\left(\frac{y}{x}\right)=y$

Solution : $x \frac{d y}{d x}+x \cos ^{2}\left(\frac{y}{x}\right)=y$

\begin{aligned} &\frac{d y}{d x}+\cos ^{2}\left(\frac{y}{x}\right)=\frac{y}{x} \\ &\frac{d y}{d x}=\frac{y}{x}-\cos ^{2}\left(\frac{y}{x}\right) \end{aligned}

Putting $y=vx$ and differentiate

Therefore, $\frac{d y}{d x}=v+x \frac{d v}{d x}$

Therefore,

\begin{aligned} &v+x \frac{d v}{d x}=v-\cos ^{2}(v) \\ &\frac{1}{\cos ^{2} v} d v=-\frac{d x}{x} \\ &\sec ^{2} v d v=-\frac{1}{x} d x \end{aligned}

Integrating both sides,

\begin{aligned} \int \sec ^{2} d v &=-\int \frac{1}{x} d x \\ \tan v &=-\log |x|+c \\ \tan \frac{y}{x} &=-\log |x|+c \end{aligned}

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