• # question_answer Let $y=y(x)$ be the solution of the differential equation $\frac{dy}{dx}+2y=f(x)$, where $f(x)=\left\{ \begin{matrix} 1, & x\in [0,1] \\ 0, & otherwise \\ \end{matrix} \right.$ If $\text{y(0)=0,then}\,\,\,\text{y}\,\,\left( \frac{\text{3}}{\text{2}} \right)\text{is}$                          [JEE Online 15-04-2018] A) $\frac{{{e}^{2}}-1}{2{{e}^{3}}}$       B) $\frac{{{e}^{2}}-1}{{{e}^{3}}}$         C) $\frac{1}{2e}$                                  D) $\frac{{{e}^{2}}+1}{2{{e}^{4}}}$

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,$y(1)=\frac{1-{{e}^{-2}}}{2}=\frac{{{e}^{2}}-1}{2{{e}^{2}}}.......(1)$ 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 (1) , 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 $y=\frac{{{e}^{2}}-1}{2{{e}^{3}}}$. So, the correct answer is option A. You will be redirected in 3 sec 