Question

The solution of $\frac{dy}{dx}=\frac{x{\mathrm{log\; x}}^2+x}{\sin y+y\cos y}$

• A. $y\sin y=x^2\log x+c$
• B. $y\sin y=x^2+c$
• C. $y\sin y=x^2+\log x+c$
• D. $y\sin y=x\log x+c$

Answer 1

Answer: (A)

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

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

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

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

Integrating both sides, we get

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

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

Consider $\int y\cos ydy$

Let $u=y\Rightarrow du=dy$

$dv=\cos ydy\Rightarrow v=\sin y$

Consider $\int x\log xdx$

Let $u=\log x\Rightarrow du=\frac{1}{x}dx$

$dv=xdx\Rightarrow v=\frac{x^2}{2}$

$\Rightarrow -\cos y+y\sin y-\int \sin ydy=2[\frac{x^2\log x}{2}-\int \frac{x^2}{2}\times \frac{1}{x}dx]+\frac{x^2}{2}+c$

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

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

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