Question

Let $y=y(x)$ be the solution of the differential equation $\frac{dy}{dx}+2y=f(x)$, where $f(x)=\{\begin{array}{ll}1& ,\text{x }ϵ[0,1]\\ 0& ,\text{otherwise }\end{array}$ If $y(0)=0$, 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

Answer: (D)

$\frac{dy}{dx}+2y=f\left(x\right)$ is a linear differential equation
If $=e^{{\int }^{2dx}}=e^{2x}$
solution of the above equation is
$y.e^{2x}={\int }_0^{x}f\left(x\right).e^{2x}dx+C$
$y\left(x\right)=e^{-2x}{\int }_0^{x}f\left(x\right)e^{2x}dx+c{e}^{-2x}$
$y\left(x\right)=0\Rightarrow c=0$
$\Rightarrow y\left(x\right)=e^{-2x}{\int }_0^{x}f\left(x\right)e^{2x}dx$
$y\left(3/2\right)=e^{-3}\left[{\int }_0^1{e}^{2x}dx+{\int }_1^{3/2}0.0x\right]=\frac{e^{-3}}{2}\left[e^2-1\right]=\frac{e^2-1}{2e^3}$