Question

General solution of differential equation $\frac{dy}{dx}+y=1(y\ne 1)$ 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,
$\frac{dy}{dx}+y=1$
$\Rightarrow \frac{dy}{dx}=1-y$
$\Rightarrow \frac{1}{1-y}dy=dx$
On integrating both the sides, we have
$\int \frac{1}{1-y}dy=\int dx$
$\Rightarrow -\log |1-y|=x+C$
$\Rightarrow \log \left|\frac{1}{1-y}\right|=x+C$