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Question

If y=y(x) is the solution of the differential equation ,eydydx1=exsuch 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


Explanation for the correct option:

Step 1: Simplifying differential equation:

The given differential equations are eydydx1=ex...i

y(0)=0...ii

From i,differentiating on both sides,

dydx1=exey=ex-y...iii

Assuming xy=z

Differentiating on both sides, with respect to x

1dydx=dzdxdydx1=dzdx...iv

From iii and iv,we get

dzdx=ez1ezdz=dxezdz=dx...v

Integrating the equation v

ez=x+ce(xy)=x+c[z=xy]

Step 2: Find the value of c:

Substituting x=0 and y(0)=0

e(0y(0))=0+ce(00)=0+ce0=cc=1

e(xy)=x+1

Step 3: Finding the value of y(1).

Taking log on both sides,

loge(xy)=logx+1xy=logx+1[loge=1]x+y=logx+1y=x+logx+1

Place it x=1now,

y(1)=1+log(1+1)

y(1)=1+loge2

Hence, the correct option is (D).


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