JEE Main 2018 (Online) 15th April Morning ShiftQuestion 1
Let $y=y(x)$ be the solution of the differential equation $\frac{dy}{dx}+2y=f(x)$, where $f(x)=\begin{cases}1,&x\in[0,1]\\0,&\text{otherwise}\end{cases}$. If $y(0)=0$, then $y\left(\frac32\right)$ is
Differential Equation
Solution
Concept. Solve on $[0,1]$ with $f=1$, then continue for $x>1$ with $f=0$, starting from $y(1)$. Formulas used. (1) $y'+2y=1\Rightarrow y=\dfrac12+Ce^{-2x}$; $\ y'+2y=0\Rightarrow y=Ke^{-2x}$Step 1 - On $[0,1]$. By (1) with $y(0)=0$: $y=\dfrac{1-e^{-2x}}{2}$, so $y(1)=\dfrac{1-e^{-2}}{2}$. Step 2 - For $x>1$. By (1), $y=y(1)\,e^{-2(x-1)}$. Step 3 - At $x=\frac32$. $$y\left(\frac32\right)=\frac{1-e^{-2}}{2}\,e^{-1}=\frac{e^2-1}{2e^3}.$$ Answer: (A)