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)

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