Question

Solve the following differential equation:

${\displaystyle [x+y\cos \left(\frac{y}{x}\right)]\text{d}x=x\cos \left(\frac{y}{x}\right)\text{d}y}$

Answer 1

The given equation can be written as

${\displaystyle \frac{\text{d}y}{\text{d}x}=\frac{x+y\cos \left(\frac{y}{x}\right)}{x\frac{\cos y}{x}}}$  ........(1)

This is a homogeneous differential equations.

Put y = vx

${\displaystyle \frac{\text{d}y}{\text{d}x}=\text{v}+x\frac{\text{dv}}{\text{d}x}}$

1 ⇒ ∴ ${\displaystyle \text{v}+x\frac{\text{dv}}{\text{d}x}=\frac{x+\text{v}x\cos \left(\frac{vx}{x}\right)}{x\cos \left(\frac{vx}{x}\right)}}$

${\displaystyle \text{v}+x\frac{\text{dv}}{\text{d}x}=\frac{x[1+\text{v}\cos \text{v}]}{\cos \text{v}}}$

${\displaystyle x\frac{\text{dv}}{\text{d}x}=\frac{1+\text{v}\cos \text{v}}{\cos \text{v}}-\text{v}}$

${\displaystyle x\frac{\text{dv}}{\text{d}x}=\frac{1}{\cos \text{v}}}$

cos v dv = ${\displaystyle \frac{\text{d}x}{x}}$

On integration we obtain

${\displaystyle \int \cos \text{v} \text{d}v=\int \frac{\text{d}x}{x}}$

sin v = log x + log c

${\displaystyle \sin \left(\frac{y}{x}\right)}$ = log x + log c

${\displaystyle \sin \left(\frac{y}{x}\right)}$ = log |cx|

Which gives the required solution.