Question

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

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

Answer 1

Answer: (A)

$\frac{e^2-1}{2e^3}$

Solving the initial value problem, we get $y=\frac{1}{2}-\frac{1}{2}{e}^{-2x}$ when $x\in [0,1]$. We can check this by substituting this in the differential equation and checking the initial value.

So, $y\left(1\right)=\frac{1-e^{-2}}{2}=\frac{e^2-1}{2e^2}...\left(1\right)$

Now, for $x\in (1,\infty )$, we have $e^{2x}y=c_2$ (solving the differential equation separately for this interval)

Using the condition found above in $\left(1\right)$, we have $c_2=\frac{e^2-1}{2}$. That gives $y=\frac{e^2-1}{2}{e}^{-2x}$ for $x\in (1,\infty )$

So, for $x=\frac{3}{2}$, we get $y=\frac{e^2-1}{2e^3}$.

So, the correct answer is option A.