Question

Find the solution of differential equation
$\frac{dy}{dx}=\frac{(\sin y+e^x)}{(\ln y-x\cos y)}$ is

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

Answer 1

The correct option is A $y\left(\right(\ln y)-1)=e^x+x\sin y+c$

$\frac{dy}{dx}=\frac{(\sin y+e^x)}{(\ln y-x\cos y)}\Rightarrow \ln ydy-x\cos ydy=\sin ydx+e^{x}dx$
Integrating both side
$\Rightarrow \int \ln ydy-\int x\cos ydy=\int \sin ydx+\int e^{x}dx$
$\Rightarrow y\ln y-y=e^x+\int d\left(x\sin y\right)+c$
$\Rightarrow y\left(\right(\ln y)-1)=e^x+x\sin y+c$