Question

General solution of differential equatin $\frac{dy}{dx}+y=1(y\ne 1)$ 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)

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

$\frac{dy}{dx}+y=1$
$\frac{dy}{dx}=1-y$
$dy=(1-y)dx$
$\frac{dy}{1-y}=dx$
Integrate both sides
$\int \frac{dy}{1-y}=\int dx$
$-\log |1-y|=X+C$
$\log {|1-y|}^{-1}=X+C$
$\log \left|\frac{1}{1-y}\right|=X+C$