Question

Let $y=y\left(x\right)$ be the solution of the differential equation $\frac{dy}{dx}+2y=f\left(x\right),$ where $f\left(x\right)=\{\begin{array}{c}1,x\in [0,1]\\ 0,\text{otherwise}\end{array}$
If $y\left(0\right)=0$, then $y\left(\frac{3}{2}\right)$ is

• A. $\frac{e^2+1}{2e^4}$
• B. $\frac{1}{2e}$
• C. $\frac{e^2-1}{e^3}$
• D. $\frac{e^2-1}{2e^3}$

Answer 1

The correct option is D $\frac{e^2-1}{2e^3}$

$\frac{dy}{dx}+2y=f\left(x\right)$
Integrating factor,
$I.F.=e^{\int 2dx}=e^{2x}$
$\Rightarrow y\cdot e^{2x}=\int e^{2x}f\left(x\right)dx+C\Rightarrow y=e^{-2x}\int e^{2x}f\left(x\right)dx+C{e}^{-2x}$
As $y\left(0\right)=0$
$\Rightarrow y\left(x\right)=e^{-2x}\underset{0}{\overset{x}{\int }}{e}^{2x}f\left(x\right)dx\Rightarrow y\left(\frac{3}{2}\right)=e^{-3}\underset{0}{\overset{3/2}{\int }}{e}^{2x}f\left(x\right)dx\Rightarrow y\left(\frac{3}{2}\right)=e^{-3}\underset{0}{\overset{1}{\int }}{e}^{2x}\cdot 1dx+e^{-3}\underset{1}{\overset{3/2}{\int }}{e}^{2x}\cdot 0dx\Rightarrow y\left(\frac{3}{2}\right)=\frac{[e^{2x}{]}_0^1}{2e^3}=\frac{e^2-1}{2e^3}$