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Exercise 6.4 · Q57

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

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y2dx+(xy+x2)dy=0y^2dx+(xy+x^2)dy=0 gives dydx=−y2x(x+y)\dfrac{dy}{dx}=-\dfrac{y^2}{x(x+y)}, homogeneous. Put y=vxy=vx: v+xdvdx=−v21+vv+x\dfrac{dv}{dx}=-\dfrac{v^2}{1+v}, so xdvdx=−v21+v−v=−v(2v+1)1+vx\dfrac{dv}{dx}=-\dfrac{v^2}{1+v}-v=-\dfrac{v(2v+1)}{1+v}, i.e. v+1v(2v+1)dv=−dxx\dfrac{v+1}{v(2v+1)}dv=-\dfrac{dx}{x}. By partial fractions v+1v(2v+1)=1v−12v+1\dfrac{v+1}{v(2v+1)}=\dfrac{1}{v}-\dfrac{1}{2v+1}, so integrating gives $\log v-\tfrac1 …

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