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Exercise 6.4 · Q59

Q.Solve: dydx+x−2y2x−y=0\dfrac{dy}{dx}+\dfrac{x-2y}{2x-y}=0

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dydx+x−2y2x−y=0\dfrac{dy}{dx}+\dfrac{x-2y}{2x-y}=0 gives dydx=2y−x2x−y\dfrac{dy}{dx}=\dfrac{2y-x}{2x-y}, homogeneous. Put y=vxy=vx: v+xdvdx=2v−12−vv+x\dfrac{dv}{dx}=\dfrac{2v-1}{2-v}, so xdvdx=v2−12−vx\dfrac{dv}{dx}=\dfrac{v^2-1}{2-v}, i.e. 2−v(v−1)(v+1)dv=dxx\dfrac{2-v}{(v-1)(v+1)}dv=\dfrac{dx}{x}. Partial fractions give 2−vv2−1=1/2v−1−3/2v+1\dfrac{2-v}{v^2-1}=\dfrac{1/2}{v-1}-\dfrac{3/2}{v+1}, so integrating: $\tfrac12\log(v-1)-\tfr …

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