Question

The general solution of the differential equation $\frac{dy}{dx}=\frac{x(2\ln x+1)}{\sin y+y\cos y},x>0$ is
(where $c$ is constant of integration)

• A. $y\sin y=x^2\ln x+c$
• B. $y\sin y=x^2(\ln x+1)+c$
• C. $y\sin y-\cos y=x^2(\ln x+1)+c$
• D. $y\sin y=(\ln x+1)+c$

Answer 1

The correct option is A $y\sin y=x^2\ln x+c$

Given : $\frac{dy}{dx}=\frac{x(2\ln x+1)}{\sin y+y\cos y}$
$\Rightarrow (\sin y+y\cos y)dy=x(2\ln x+1)dx$
Integrating both sides, we get
$\int (\sin y+y\cos y)dy=\int x(2\ln x+1)dx$
$\int \left(\sin y\right)dy+\int \left(y\cos y\right)dy=\int x(2\ln x+1)dx$
Using $ILATE$ rule, we get
$-\cos y+y\sin y+\cos y=x^2\ln x+\frac{x^2}{2}-\frac{x^2}{2}+c$
$\Rightarrow y\sin y=x^2\ln x+c$