Let y=y(x) be the solution of the differential equation d y d x+2 y=f(x), where f(x)= … . If y(0)=0, then y ( 3 2 ) is

When $x \in[0,1]$x∈[0,1], then $\frac{d y}{d x}+2 y=1$dxdy​+2y=1 $$\begin{aligned} &\Rightarrow \mathrm{y}=\frac{1}{2}+C_1 e^{-2 x} \\ &\because y(0)=0 \Rightarrow y(x)=\frac{1}{2}-\frac{1}{2} e^{-2 x} \\ &\text { Here, } y(1)=\frac{1}{2}-\frac{1}{2} e^{-2}=\frac{e^2-1}{2 e^2} \end{aligned}$$​⇒y=21​+C1​e−2x∵y(0)=0⇒y(x)=21​−21​e−2x Here, y(1)=21​−21​e−2=2e2e2−1​​ When $x \notin[0,1]$x∈/[0,1], than $\frac{d y}{d x}+2 y=0$dxdy​+2y=0 $$\begin{gathered} \Rightarrow y=c_2 e^{-2 x} \\ \therefore y(1)=\frac{e^2-1}{2 e^2} \Rightarrow \frac{e^2-1}{2}=c^2 e^{-2} \Rightarrow C_2=\frac{e^2-1}{2} \\ \therefore y(x)\left(\frac{e^2-1}{2}\right) e^{-2 x} \Rightarrow y\left(\frac{3}{2}\right)=\frac{e^2-1}{2 e^3} \end{gathered}$$⇒y=c2​e−2x∴y(1)=2e2e2−1​⇒2e2−1​=c2e−2⇒C2​=2e2−1​∴y(x)(2e2−1​)e−2x⇒y(23​)=2e3e2−1​​