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Exercise 6.4 · Q58

Q.Solve: (x2−y2)dx+2xy⋅dy=0(x^2-y^2)dx+2xy\cdot dy=0

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(x2−y2)dx+2xy dy=0(x^2-y^2)dx+2xy\,dy=0 gives dydx=y2−x22xy\dfrac{dy}{dx}=\dfrac{y^2-x^2}{2xy}, homogeneous. Put y=vxy=vx: v+xdvdx=v2−12vv+x\dfrac{dv}{dx}=\dfrac{v^2-1}{2v}, so xdvdx=−v2+12vx\dfrac{dv}{dx}=-\dfrac{v^2+1}{2v}, i.e. 2vv2+1dv=−dxx\dfrac{2v}{v^2+1}dv=-\dfrac{dx}{x}. Integrating: log⁡(v2+1)=−log⁡x+c1\log(v^2+1)=-\log x+c_1, i.e. $x( …

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