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Question

Let y=y(x) be the solution of the differential equation dydx+2y=f(x), where f(x)={1 , x[0,1]0 , otherwise
If y(0)=0, then y(32) is

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

The correct option is D e212e3
dydx+2y=f(x)
Integrating factor,
I.F.=e2 dx=e2x
ye2x=e2xf(x) dx+Cy=e2xe2xf(x) dx+Ce2x
As y(0)=0
y(x)=e2xx0e2xf(x) dxy(32)=e33/20e2xf(x) dxy(32)=e310e2x1 dx+e33/21e2x0 dxy(32)=[e2x]102e3=e212e3

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