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

Answer 1

The correct option is A $\log \left|\frac{1}{1-y}\right|=x+C$

$\frac{dy}{dx}+y=1$

$\Rightarrow \frac{dy}{dx}=1-y$

$\Rightarrow \frac{dy}{1-y}=dx$

$\Rightarrow \int \frac{dy}{1-y}$$=\int dx$

$\Rightarrow -\log |1-y|=x+C$

$\Rightarrow \log \left|\frac{1}{1-y}\right|=x+C$