Question

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

Answer 1

Seperating terms of $y$ and $x$

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

By integrating both sides, we get

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

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

$-cosy+\int y\cos ydy=2\int x\log xdx+\frac{x^2}{2}$ ......$\left(1\right)$

Consider $\int y\cos ydy$

Let $u=y\Rightarrow du=dy$ and $dv=\cos ydy\Rightarrow v=\sin y$

We have $\int u.dv=uv-\int v.du$

$\int y\cos ydy=y\sin y-\int \sin ydy$

$\int y\cos ydy=y\sin y-(-\cos y)$

$\int y\cos ydy=y\sin y+\cos y$ ......$\left(2\right)$

Consider $\int x\log xdx$

Let $u=\log x\Rightarrow du=\frac{1}{x}dx$ and $dv=x.dx\Rightarrow v=\frac{x^2}{2}$

We have $\int u.dv=uv-\int v.du$

$\Rightarrow \int x\log xdx=\log x\frac{x^2}{2}-\int \frac{x^2}{2}\times \frac{1}{x}dx$

$\Rightarrow \int x\log xdx=\frac{x^2\log x}{2}-\int \frac{x}{2}dx$

$\Rightarrow \int x\log xdx=\frac{x^2\log x}{2}-\frac{x^2}{4}$ ......$\left(3\right)$

Substituting $\left(2\right)$ and $\left(3\right)$ in $\left(1\right)$ we get

$-cosy+\int y\cos ydy=2\int x\log xdx+\frac{x^2}{2}$

$-\cos y+y\sin y+\cos y=2[\frac{x^2\log x}{2}-\frac{x^2}{4}]+\frac{x^2}{2}+c$ where $c$ is the constant of integration.

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

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

At $x=1,y=\frac{\pi }{2}$ we have

$\Rightarrow \frac{\pi }{2}\sin \frac{\pi }{2}=1^2\log 1+c$ where $\log 1=0$

$\Rightarrow c=\frac{\pi }{2}$ where $\sin \frac{\pi }{2}=1$

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