Question

The solution of the equation $\frac{dy}{dx}=\frac{x(2\log x+1)}{\sin y+y\cos y}$ is

• A. $y\sin y=x^2\log x+\frac{x^2}{y}+c$
• B. $y\cos y=x^2(\log x+1)+c$
• C. $y\cos y=x^2\log x+\frac{x^2}{2}+c$
• D. $y\sin y=x^2\log x+c$

Answer 1

The correct option is D $y\sin y=x^2\log x+c$

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

Therefore, $(y\cos y+\sin y)dy=(2x\log x+x)dx$
$\Rightarrow y\sin y-\int \sin ydy+\int \sin ydx=x^2\log x-\int x^2\cdot \frac{1}{x}dx+\int xdx+c$
$\Rightarrow y\sin y=x^2\log x+c$