Question

Solve the following differential equation : \[\left[ y - x  \cos\left( \frac{y}{x} \right) \right]dy + \left[ y  \cos\left( \frac{y}{x} \right) - 2x  \sin\left( \frac{y}{x} \right) \right]dx = 0\] .

Answer 1

The given differential equation is

\[\left[ y - x \cos\left( \frac{y}{x} \right) \right]dy + \left[ y \cos\left( \frac{y}{x} \right) - 2x \sin\left( \frac{y}{x} \right) \right]dx = 0\].

\[\therefore \frac{d y}{d x} = \frac{2x\sin\left( \frac{y}{x} \right) - y\cos\left( \frac{y}{x} \right)}{y - x\cos\left( \frac{y}{x} \right)}\]

This is a homogeneous differential equation.
Putting y = vx and \[\frac{d y}{d x} = v + x\frac{d v}{d x}\] it reduces to

\[v + x\frac{d v}{d x} = \frac{2x\sin v - vx\cos v}{vx - x\cos v}\]

\[ \Rightarrow v + x\frac{d v}{d x} = \frac{2\sin v - v\cos v}{v - \cos v}\]

\[ \Rightarrow x\frac{d v}{d x} = \frac{2\sin v - v\cos v}{v - \cos v} - v\]

\[ \Rightarrow x\frac{d v}{d x} = \frac{2\sin v - v\cos v - v^2 + v\cos v}{v - \cos v}\]

\[ \Rightarrow x\frac{d v}{d x} = \frac{2\sin v - v^2}{v - \cos v}\]

\[ \Rightarrow \left( \frac{v - \cos v}{2\sin v - v^2} \right)dv = \frac{dx}{x}\]

\[ \Rightarrow \frac{- 1}{2}\left( \frac{2\cos v - 2v}{2\sin v - v^2} \right)dv = \frac{dx}{x}\]

Integrating on both sides, we get

\[- \frac{1}{2}\int\frac{2\cos v - 2v}{2\sin v - v^2}dv = \int\frac{dx}{x}\]

\[ \Rightarrow - \frac{1}{2}\log\left( 2\sin v - v^2 \right) = \log x + \log C \left( \int\frac{f'\left( x \right)}{f\left( x \right)}dx = \log x + C \right)\]

\[\text { where} \]

\[C =\text { Constant of integration } \]

\[\Rightarrow - \frac{1}{2}\log\left( 2\sin v - v^2 \right) = \log x + \log C\]

\[ \Rightarrow \log\left( \frac{1}{2\sin v - v^2} \right) = 2\log Cx\]

\[ \Rightarrow \frac{1}{2\sin v - v^2} = C^2 x^2 \]

\[ \Rightarrow x^2 \left[ 2\sin\left( \frac{y}{x} \right) - \frac{y^2}{x^2} \right] = \frac{1}{C^2} = k\]

\[ \Rightarrow 2 x^2 \sin\left( \frac{y}{x} \right) - y^2 = k\]

This is the solution of the given differential equation.