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Question

Let $$y = y(x)$$ be the solution of the differential equation $$\dfrac {dy}{dx} + 2y = f(x)$$, where $$f(x) = \left\{\begin{matrix} 1, & x\in [0, 1]\\ 0, & otherwise\end{matrix}\right.$$
If $$y(0) = 0$$, then $$y\left (\dfrac {3}{2}\right )$$ is


A
e212e3
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B
e21e3
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C
12e
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D
e2+12e4
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Solution

The correct option is D $$\dfrac {e^{2} - 1}{2e^{3}}$$
Solving the initial value problem, we get $$y = \dfrac{1}{2} - \dfrac{1}{2}e^{-2x}$$ when $$x \in [0, 1]$$. We can check this by substituting this in the differential equation and checking the initial value.
So, $$y(1) = \dfrac{1-e^{-2}}{2} = \dfrac{e^2-1}{2e^2} ... (1)$$
Now, for $$x \in (1, \infty)$$, we have $$e^{2x}y = c_2$$ (solving the differential equation separately for this interval)
Using the condition found above in $$(1)$$, we have $$c_2 = \dfrac{e^2-1}{2}$$. That gives $$y = \dfrac{e^2-1}{2}e^{-2x}$$ for $$x \in (1, \infty)$$
So, for $$x=\dfrac{3}{2}$$, we get $$y = \dfrac{e^2-1}{2e^3}$$. 
So, the correct answer is option A.

Mathematics

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