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Question

If y=y(x) is the solution curve of the differential equation x2dy+(y1x)dx=0; x>0, and y(1)=1, then y(12) is equal to:

A
3e
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B
321e
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C
3+1e
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D
3+e
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Solution

The correct option is A 3e
x2dy=(1xy)dx
x2dydx=1xy
x2dydx+y=1x
dydx+1x2.y=1x3(i)
IF=e1x2dx=e1x
Therefore, the solution is
y.e1x=e1x×1x3dx
Let 1x=t, 1x2dx=dt
y.e1x=e(t)tdt=et(t1)+c
ye1x=e1x(1x1)+c
y=1x+1+ce1x
y(1)=11=2+cec=1e
y(12)=2+1+(1e).e2
=3e

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