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Question

For the primitive integral equation ydx+y2dy=xdy;xR,y>0,y(x),y(1)=1, then y(3) is:

A
3
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B
2
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C
1
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D
5
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Solution

The correct option is B 1
ydx+y2dy=xdy
ydxxdyy2=dy
xy=y+c
=>y(1)=1c=2y2+2y+x=0
=>y(3):y2+2y3=0(y+3)(y1)=0=>y=3,1

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