Question

What is the solution of $\frac{dy}{dx}+2y=1$ satisfying $y(0)=0$ ?

• A. $y=\frac{1-e^{-2x}}{2}$
• B. $y=\frac{1+e^{-2x}}{2}$
• C. $y=1+e^x$
• D. $y=\frac{1+e^x}{2}$

Answer 1

Answer: (A)

$\frac{dy}{dx}+2y=1$
$\Rightarrow \frac{dy}{dx}=1-2y$
$\int \frac{dy}{1-2y}=\int dx-\frac{1}{2}\log |1-2y|=x+C$
at $x=0,y=0;-\frac{1}{2}\log 1=0+C$
$\Rightarrow C=0$
$1-2y=e^{-2x}$
$\Rightarrow y=\frac{1-e^{-2x}}{2}$