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$The solution of primitive integral equation (x^{2} + y^{2}) dy = xydx is y = y (x) . If$ $y (1) = 1 and y (x_{0}) = e then x_{0} is$

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a

2(e2−1)

b

2(e2+1)

c

3 e

d

e2+12

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detailed solution

Correct option is C

Put y=Vx ⇒ dydx=V+xdvdxfrom given dydx=xyx2+y2V+xdvdx=Vx2x21+V2=V1+V2xdvdx=V−V−V31+V2=−V31+V21+V2V3dv=-dxx→∫1V3+V2V3dv=−∫dxx→−12V2+log⁡V=−log⁡x+C→−x22y2+log⁡y−log⁡x=−log⁡x+C→log⁡y−x22y2=CPasses through ⁡(1,1)_log1-12=C, C=−12⇒log⁡y−x22e2=−12 Put y=e⇒loge −x22e2=−121+12=x22e2⇒x22e2=32⇒x2=3e2⇒x02=3e2⇒x0=3e

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