Question

The general solution of
$\sin y \cdot \frac{d y}{d x}=\cos y(1-x \cos y)$ 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{d y}{d x}=\cos y(1-x \cos y) $
$ \Rightarrow \sin y \frac{d y}{d x}=\cos y-x \cos ^{2} y$
$ \Rightarrow \frac{\sin y}{\cos ^{2} y} \frac{d y}{d x}=\frac{1}{\cos y}-x $
$\Rightarrow \sec y \tan y \frac{d y}{d x}=\sec y-x $
Let $\sec y=t$
$\Rightarrow \sec y \tan y \frac{d y}{d x}=\frac{d t}{d x}$
$ \therefore \frac{d t}{d x} =t-x $
$ \Rightarrow \frac{d t}{d x}+ (-t) =-x $
$ \therefore IF =e^{\int-d x}=e^{-x}$
Now, required solution
$ \Rightarrow t(I F) =\int(-x)(I F) d x+c $
$ \sec y\left(e^{-x}\right) =\int(-x) e^{-x} d x+c $
$=\left(x e^{-x}\right)-\int e^{-x} d x+c $
$=x e^{-x}+e^{-x}+c $
$ \Rightarrow \sec y=x+1+c e^{x} $