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Question

If y=y(x) is the solution of the differential equation, ey(dydx1)=ex such that y(0)=0, then y(1) is equal to

A
loge2
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B
2e
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C
2+loge2
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D
1+loge2
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Solution

The correct option is D 1+loge2
ey(dydx1)=ex
dydx=exy+1

Let xy=t
1dydx=dtdx
So, we can write
1dtdx=et+1
etdt=dx
et=x+c
eyx=x+c

y(0)=0
1=0+cc=1
eyx=x+1

At x=1,
ey1=2
y=1+loge2

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