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Exercise 6.4 · Q63

Q.Solve: xydydx=x2+2y2xy\dfrac{dy}{dx}=x^2+2y^2, y(1)=0y(1)=0

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xydydx=x2+2y2xy\dfrac{dy}{dx}=x^2+2y^2 gives dydx=1+2v2v\dfrac{dy}{dx}=\dfrac{1+2v^2}{v} under y=vxy=vx: v+xdvdx=1+2v2vv+x\dfrac{dv}{dx}=\dfrac{1+2v^2}{v}, so xdvdx=1+v2vx\dfrac{dv}{dx}=\dfrac{1+v^2}{v}, i.e. v1+v2dv=dxx\dfrac{v}{1+v^2}dv=\dfrac{dx}{x}. Integrating: 12log⁡(1+v2)=log⁡x+c1\tfrac12\log(1+v^2)=\log x+c_1, i.e. 1+v2x2=c\dfrac{1+v^2}{x^2}=c. Substitut …

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