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Question:
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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
$$
f(x)=\left\{\begin{array}{lc}
1, & x \in[0,1] \\
0, & \text { otherwise }
\end{array}\right.
$$
If $y(0)=0$, then $y\left(\frac{3}{2}\right)$ is
Options:
$$
f(x)=\left\{\begin{array}{lc}
1, & x \in[0,1] \\
0, & \text { otherwise }
\end{array}\right.
$$
If $y(0)=0$, then $y\left(\frac{3}{2}\right)$ is
Solution:
1213 Upvotes
Verified Answer
The correct answer is:
$\frac{e^2-1}{2 e^3}$
$\frac{e^2-1}{2 e^3}$
When $x \in[0,1]$, then $\frac{d y}{d x}+2 y=1$
$$
\begin{aligned}
&\Rightarrow \mathrm{y}=\frac{1}{2}+C_1 e^{-2 x} \\
&\because y(0)=0 \Rightarrow y(x)=\frac{1}{2}-\frac{1}{2} e^{-2 x} \\
&\text { Here, } y(1)=\frac{1}{2}-\frac{1}{2} e^{-2}=\frac{e^2-1}{2 e^2}
\end{aligned}
$$
When $x \notin[0,1]$, than $\frac{d y}{d x}+2 y=0$
$$
\begin{gathered}
\Rightarrow y=c_2 e^{-2 x} \\
\therefore y(1)=\frac{e^2-1}{2 e^2} \Rightarrow \frac{e^2-1}{2}=c^2 e^{-2} \Rightarrow C_2=\frac{e^2-1}{2} \\
\therefore y(x)\left(\frac{e^2-1}{2}\right) e^{-2 x} \Rightarrow y\left(\frac{3}{2}\right)=\frac{e^2-1}{2 e^3}
\end{gathered}
$$
$$
\begin{aligned}
&\Rightarrow \mathrm{y}=\frac{1}{2}+C_1 e^{-2 x} \\
&\because y(0)=0 \Rightarrow y(x)=\frac{1}{2}-\frac{1}{2} e^{-2 x} \\
&\text { Here, } y(1)=\frac{1}{2}-\frac{1}{2} e^{-2}=\frac{e^2-1}{2 e^2}
\end{aligned}
$$
When $x \notin[0,1]$, than $\frac{d y}{d x}+2 y=0$
$$
\begin{gathered}
\Rightarrow y=c_2 e^{-2 x} \\
\therefore y(1)=\frac{e^2-1}{2 e^2} \Rightarrow \frac{e^2-1}{2}=c^2 e^{-2} \Rightarrow C_2=\frac{e^2-1}{2} \\
\therefore y(x)\left(\frac{e^2-1}{2}\right) e^{-2 x} \Rightarrow y\left(\frac{3}{2}\right)=\frac{e^2-1}{2 e^3}
\end{gathered}
$$
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