Question

General solution of differential equation $\frac{dy}{dx} + y=1 (y \neq 1)$ is

• A. $\log \left| \frac{1}{1-y}\right| = x + C$
• B. $\log | 1 - y| = x + C$
• C. $\log | 1 + y| = x + C$
• D. $\log \left| \frac{1}{1-y}\right| = - x + C$

Answer 1

Answer: (A)

We have,
$ \frac{d y}{d x}+y=1 $
$\Rightarrow \frac{d y}{d x}=1-y $
$\Rightarrow \frac{1}{1-y} d y=d x$
On integrating both the sides, we have
$\int \frac{1}{1-y} d y=\int \, dx$
$\Rightarrow \,-\log |1-y|=x+C$
$\Rightarrow \, \log \left|\frac{1}{1-y}\right|=x+C$