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$For the primitive integral equation ydx + y^{2} dy = xdy, x \in R, y > 0 y = y (x), y (1) = 1 then y (- 3) is$

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Correct option is A

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

Remember concepts with our Masterclasses.

Ready to Test Your Skills?

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$For the primitive integral equation ydx + y^{2} dy = xdy, x \in R, y > 0 y = y (x), y (1) = 1 then y (- 3) is$