Question

Find the particular solution of the differential equation $\frac{dy}{dx}=\frac{x(2llogx+1)}{siny+ycosy}$ given that
$y=\frac{\pi }{2}$ where x = 1 .

Answer 1

$\frac{dy}{dx}=\frac{x(2\log x+1)}{\sin y+\cos y+y}$

Separating,

$\Rightarrow (\sin y+y\cos y)dy=(2x\log x+x)dx$

Interating both sides

$\Rightarrow -\cos y+y\sin y+\cos y=2x^2\log \frac{x}{2}-\frac{2x^2}{4}+\frac{x^2}{2}+cy\sin y$

$=x^2\log x-\frac{x^2}{2}+\frac{x^2}{2}+c$

$\therefore y\sin y=x^2\log x+c$

Putting $x=1$ $\&$ $y=90,$ $c=\frac{\pi }{2}$

Required answer $y\sin y=x^2\log x+\frac{\pi }{2}$

Hence, the answer is $y\sin y=x^2\log x+\frac{\pi }{2}.$