If $\cos \frac{y}{x}=A \log x+C$ is the general solution of $\left(x \sin \frac{y}{x}\right) d y=\left(y…
If $\cos \frac{y}{x}=A \log x+C$ is the general solution of $\left(x \sin \frac{y}{x}\right) d y=\left(y \sin \frac{y}{x}-x\right) d x$, then $A=$
- 2
- 1
- -1
- -2
Solution
$
\begin{aligned}
& \text { }\left(x \sin \frac{y}{x}\right) d y=\left(y \sin \frac{y}{x}-x\right) d x \\
& \frac{d y}{d x}=\frac{y \sin y / x-x}{x \sin \frac{y}{x}} \Rightarrow \frac{d y}{d x}=\frac{y}{x}-\frac{1}{\sin \frac{y}{x}} \\
& \text { Let } \frac{y}{x}=v \text { or } y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x} \\
& \Rightarrow v+x \frac{d v}{d x}=v-\frac{1}{\sin v} \Rightarrow-\int \sin v d v=\int \frac{1}{x} d x \\
& \Rightarrow \cos v=\log x+C \Rightarrow \cos \frac{y}{x}=\log x+C
\end{aligned}
$
Compare it with given equation, we get $A=1$
Asked in: AP EAMCET 2022 (06 Jul Shift 2)
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