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Question

Consider the differential equation dydx+y=ex with y(0) = 1. Then the value of y(1) is

A
e + e1
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B
12[ee1]
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C
12[e+e1]
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D
2[ee1]
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Solution

The correct option is C 12[e+e1]
dydx+y=ex ...(1)

It is LDE in y & x

I.F = edx=ex

The general solution of equation (1) is

y(I.F) = (I.F)exdx+C

y.ex=exexdx+C

y.ex=e2x2+C...(2)

Now using y(0) = 1 C=12

Hence Solution is

yex=e2x2+12

y(1)=e2+e12=e+e12

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