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Exercise 6.4 · Q65

Q.Solve: x2dydx=x2+xy+y2x^2\dfrac{dy}{dx}=x^2+xy+y^2

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x2dydx=x2+xy+y2x^2\dfrac{dy}{dx}=x^2+xy+y^2 gives dydx=1+v+v2\dfrac{dy}{dx}=1+v+v^2 under y=vxy=vx: v+xdvdx=1+v+v2v+x\dfrac{dv}{dx}=1+v+v^2, so xdvdx=1+v2x\dfrac{dv}{dx}=1+v^2, i.e. dv1+v2=dxx\dfrac{dv}{1+v^2}=\dfrac{dx}{x}. Integrating: tan⁡−1v=log⁡x+c\tan^{-1}v=\log x+c. Substituting v=y/xv=y/x: $\tan^{ …

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