Q.

For the primitive integral equation  ydx+y2dy=xdy, x∈R,y>0   y=y(x), y(1)=1 then y-3 is

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a

3

b

2

c

1

d

5

answer is A.

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Detailed Solution

ydx−xdyy2=−dy⇒∫dxy=−∫dy⇒xy=−y+c⇒y(1)=1⇒2=c⇒xy=−y+2⇒y2−2y+x=0 Put x=-3⇒y2−2y−3=0⇒y=3,−1
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