Question

The general solution of
$\sin y\cdot \frac{dy}{dx}=\cos y(1-x\cos y)$ is

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

Answer 1

Answer: (B)

Given differential equation is
$\sin y\frac{dy}{dx}=\cos y(1-x\cos y)$
$\Rightarrow \sin y\frac{dy}{dx}=\cos y-x{\cos }^{2}y$
$\Rightarrow \frac{\sin y}{{\cos }^{2}y}\frac{dy}{dx}=\frac{1}{\cos y}-x$
$\Rightarrow \sec y\tan y\frac{dy}{dx}=\sec y-x$
Let $\sec y=t$
$\Rightarrow \sec y\tan y\frac{dy}{dx}=\frac{dt}{dx}$
$\therefore \frac{dt}{dx}=t-x$
$\Rightarrow \frac{dt}{dx}+(-t)=-x$
$\therefore IF=e^{\int -dx}=e^{-x}$
Now, required solution
$\Rightarrow t(IF)=\int (-x)(IF)dx+c$
$\sec y\left(e^{-x}\right)=\int (-x)e^{-x}dx+c$
$=\left(x{e}^{-x}\right)-\int e^{-x}dx+c$
$=x{e}^{-x}+e^{-x}+c$
$\Rightarrow \sec y=x+1+c{e}^x$