Question

Find the particular solution of the differential equation $\frac{dy}{dx}=\frac{x(2\log x+1)}{\sin y+y\cos y}$ given that $y=\frac{\pi }{2}$ when $x=1$.

Answer 1

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

$\Rightarrow (\sin y+y\cos y)dy=x(2\log x+1)dx$

$\Rightarrow \int (\sin y+y\cos y)dy=\int x(2\log x+1)dx$

$\Rightarrow \int \sin ydy+\int y\cos ydy=\int 2x\log xdx+\int xdx$

$\Rightarrow -\cos y+y\sin y+\cos y=2\left(\frac{1}{2}{x}^2\ln \right(x)-\frac{x^2}{4})+\frac{x^2}{2}+C$

$\Rightarrow y\sin y=\left(x^2\ln \right(x)-\frac{x^2}{2})+\frac{x^2}{2}+C$

$\Rightarrow y\sin y=x^2\ln \left(x\right)+C$

The required solution is $x^2\ln \left(x\right)-y\sin y+C=0.$

When $(x,y)\equiv (1,\frac{\pi }{2})$,

We get $C=\frac{\pi }{2}$

Hence, the particular solution is $x^2\ln \left(x\right)-y\sin y+\frac{\pi }{2}=0.$