Question

Solution of $\frac{dy}{dx}=\frac{x\log x^2+x}{\sin y+y\cos y}$ is

• 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)

Given equation is $\frac{dy}{dx}=\frac{x\log x^2+x}{\sin y+y\cos y}$
$\Rightarrow (\sin y+y\cos y)dy=\left(x\log x^2+x\right)dx$
On integrating both sides, we get
$\int (\sin y+y\cos y)dy=\int \left(x\log x^2+x\right)dx$
$\Rightarrow -\cos y+y\sin y+\cos y=\frac{x^2}{2}\log x^2-\int \frac{x^2}{2}\cdot \frac{1}{x^2}2xdx+\int xdx+C$
$\Rightarrow y\sin y=\frac{x^2}{2}\cdot 2\log x-\int xdx+\int xdx+C$
$\Rightarrow y\sin y=x^2\log x+C$