Mathematics - Differential Equation Question with Solution | TestHub

When $x\in [0,1]$, then $\frac{dy}{dx}+2y=1$ $$\begin{array}{rl}& \Rightarrow \mathrm{y}=\frac{1}{2}+C_1{e}^{-2x}\\ & \because y(0)=0\Rightarrow y(x)=\frac{1}{2}-\frac{1}{2}{e}^{-2x}\\ & \text{ Here, }y(1)=\frac{1}{2}-\frac{1}{2}{e}^{-2}=\frac{e^2-1}{2e^2}\end{array}$$ When $x\notin [0,1]$, than $\frac{dy}{dx}+2y=0$ $$\begin{array}{c}\Rightarrow y=c_2{e}^{-2x}\\ \therefore y(1)=\frac{e^2-1}{2e^2}\Rightarrow \frac{e^2-1}{2}=c^2{e}^{-2}\Rightarrow C_2=\frac{e^2-1}{2}\\ \therefore y(x)\left(\frac{e^2-1}{2}\right)e^{-2x}\Rightarrow y\left(\frac{3}{2}\right)=\frac{e^2-1}{2e^3}\end{array}$$