Question

Solution of the differential equation $\sin y\cdot \frac{dy}{dx}=\cos y(1-x\cos y)$ is:
(where $C$ is integration constant)

• A. $\sec y=(1+x)e^x+C{e}^{-x}$
• B. $\sec y=(1-x)+C{e}^x$
• C. $\sec y=(1+x)+C{e}^x$
• D. $\cos y=(1+x)+C{e}^x$

Answer 1

The correct option is C $\sec y=(1+x)+C{e}^x$

The given differential equation is
$\sin y\frac{dy}{dx}=\cos y(1-x\cos y)$
$\Rightarrow \sin y\frac{dy}{dx}-\cos y=-x{\cos }^{2}y$
Dividing both sides by ${\cos }^{2}y,$ we get:
$\tan y\cdot \sec y\cdot \frac{dy}{dx}-\sec y=-x$
Put$\sec y=v\Rightarrow \sec y\tan y\cdot \frac{dy}{dx}=\frac{dv}{dx}$
So, the given differential equation converts to: $\frac{dv}{dx}-v=-x$
which is linear differential equation with $P=-1,Q=-x$
$I.F.=e^{\int Pdx}=e^{\int -1dx}=e^{-x}$

The solution is given by
$v\cdot e^{-x}=\int -x\cdot e^{-x}dx=x{e}^{-x}+e^{-x}+C$
$\Rightarrow v\cdot e^{-x}=e^{-x}(x+1)+C$
$\Rightarrow v=(1+x)+C{e}^x$
$\Rightarrow \sec y=(1+x)+C{e}^x$