Question

Find the particular solution of the differential equation x cos\[\left( \frac{y}{x} \right)\frac{dy}{dx} = y \cos\left( \frac{y}{x} \right) + x\], given that when x = 1, \[y = \frac{\pi}{4}\]

Answer 1

\[x \cos \left( \frac{y}{x} \right)\frac{dy}{dx} = y \cos \left( \frac{y}{x} \right) + x\]
\[ \Rightarrow \frac{dy}{dx} = \frac{y \cos \left( \frac{y}{x} \right) + x}{x \cos \left( \frac{y}{x} \right)}\]
This is a homogeneous differential equation .
\[\text{ Putting }y = vx\text{ and }\frac{dy}{dx} = v + x\frac{dv}{dx},\text{ we get }\]
\[v + x\frac{dv}{dx} = \frac{vx \cos v + x}{x \cos v}\]
\[ \Rightarrow v + x\frac{dv}{dx} = \frac{v \cos v + 1}{\cos v}\]
\[ \Rightarrow x\frac{dv}{dx} = \frac{v \cos v + 1 - v \cos v}{\cos v}\]
\[ \Rightarrow x\frac{dv}{dx} = \frac{1}{\cos v}\]
\[ \Rightarrow \cos v dv = \frac{1}{x}dx\]
Integrating both sides, we get
\[\int\cos v dv = \int\frac{1}{x}dx\]
\[ \Rightarrow \sin v = \log \left| x \right| + C\]
\[\text{ Putting }v = \frac{y}{x}, \text{ we get }\]
\[\sin\frac{y}{x} = \log \left| x \right| + C . . . . . \left( 1 \right)\]
\[\text{ At }x = 1, y = \frac{\pi}{4} ............\left(\text{Given} \right)\]
\[\text{ Putting }x = 1\text{ and }y = \frac{\pi}{4}\text{ in }(1),\text{ we get }\]
\[C = \frac{1}{\sqrt{2}}\]
\[\text{Putting }C = \frac{1}{\sqrt{2}}\text{ in }(1),\text{ we get }\]
\[\sin \frac{y}{x} = \log \left| x \right| + \frac{1}{\sqrt{2}}\]
\[\text{Hence, }\sin \frac{y}{x} = \log x + \frac{1}{\sqrt{2}}\text{ is the required solution .}\]