Question

Let $y = y(x)$ be the solution of the differential equation $\frac{dy}{dx} + 2y = f(x)$, where $f(x) = \begin{cases} 1 &, \text{x } \epsilon [0,1] \\ 0 &, \text{otherwise } \end{cases}$ 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 $=e^{\int^{2 dx}}=e^{2x}$
solution of the above equation is
$y.e^{2x}=\int^{x}_{0} f \left(x\right).e^{2 x}dx+C$
$y\left(x\right)=e^{-2x}\int^{x}_{0} f \left(x\right)e^{2x} dx+ce^{-2x}$
$y\left(x\right)=0 \Rightarrow c=0$
$\Rightarrow y\left(x\right)=e^{-2x} \int^{x}_{0}f\left(x\right)e^{2x} dx$
$y\left(3/2\right) =e^{-3}\left[\int^{1}_{0} e^{2x}dx+\int_{1}^{3/2} 0.0x\right]=\frac{e^{-3}}{2}\left[e^{2}-1\right]=\frac{e^{2}-1}{2e^{3}}$