Question

Solve the following differential equation :
$[y-xcos\frac{y}{x}]dy+[ycos\frac{y}{x}-2xsin\frac{y}{x}]dx=0.$

Answer 1

concept $\to$ [Extract differential equation]

$Mdx+Ndy=0$ is given difference equation

If $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}\to$ Then given equation is exact And solution is given by

$\int Mdx+\int \left(termsofNnotcontainingx\right)dy=C$

Given question

$\to \underset{\to =N}{\underset{}{[y-x\cos \frac{y}{x}]}}dy+\underset{\to =M}{\underset{}{\left[y\cos \frac{y}{x}-2x\sin \frac{y}{x}\right]}}dx=0$

So, cheching whether give difference equation is exact or not.

$\Rightarrow \frac{\partial M}{\partial y}=y[-\sin \left(\frac{y}{x}\right)]\times \frac{1}{x}+\cos \left(\frac{y}{x}\right)-2x\cos \left(\frac{y}{x}\right)\times \frac{1}{x}$

$=-\frac{y}{x}\sin \left(\frac{y}{x}\right)+\cos \left(\frac{y}{x}\right)-2\cos \left(\frac{y}{x}\right)$

$=-\frac{y}{x}\sin \left(\frac{y}{x}\right)-\cos \left(\frac{y}{x}\right)....\left(1\right)$

$\Rightarrow \frac{\partial N}{\partial x}=0-[\cos \left(\frac{y}{x}\right)+x(-\sin \left(\frac{y}{x}\right))\times \frac{y}{x^2}]$

$=-[\cos \left(\frac{y}{x}\right)+\frac{y}{x}\sin \left(\frac{y}{x}\right)]$

$=-\frac{y}{x}\sin \left(\frac{y}{x}\right)-\cos \left(\frac{y}{x}\right)....\left(2\right)$

$\Rightarrow \left(1\right)$ & $\left(2\right),\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$\to$ So, given equation is exact

$\therefore$ solution is given by ${\int }_{yconstant}Mdx+\int (termofNdy=Cnotcontaining)$

${\int }_{yconstant}[y\cos \frac{y}{x}-2x\sin \frac{y}{x}]dx+\int ydy=C$

${\int }_{yconstant}[y\cos \frac{y}{x}-2x\sin \left(\frac{y}{x}\right)dx+\frac{y^2}{2}=C]$