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

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

Aicharya · Accepted Answer

When $x \in[0,1]$, then $\frac{d y}{d x}+2 y=1$ $ $ \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} \ &	ext { Here, } y(1)=\frac{1}{2}-\frac{1}{2} e^{-2}=\frac{e^2-1}{2 e^2} \end{aligned} $ $ When $x 
otin[0,1]$, than $\frac{d y}{d x}+2 y=0$ $ $ \begin{gathered} \Rightarrow y=c_2 e^{-2 x} \ 	herefore 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} \ 	herefore y(x)\left(\frac{e^2-1}{2}ight) e^{-2 x} \Rightarrow y\left(\frac{3}{2}ight)=\frac{e^2-1}{2 e^3} \end{gathered} $ $

Let $y=y(x)$ be the solution of the differential equation $\frac{d y}{d x}+2 y=f(x)$, where $…

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