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Exercise 6.4 · Q62

Q.Solve: y2−x2dydx=xydydxy^2-x^2\dfrac{dy}{dx}=xy\dfrac{dy}{dx}

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y2−x2dydx=xydydxy^2-x^2\dfrac{dy}{dx}=xy\dfrac{dy}{dx} gives dydx=y2x2+xy=y2x(x+y)\dfrac{dy}{dx}=\dfrac{y^2}{x^2+xy}=\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=v2−v(1+v)1+v=−v1+vx\dfrac{dv}{dx}=\dfrac{v^2-v(1+v)}{1+v}=-\dfrac{v}{1+v}, i.e. 1+vvdv=−dxx\dfrac{1+v}{v}dv=-\dfrac{dx}{x}, i.e. (1v+1)dv=−dxx\left(\dfrac1v+1\right)dv=-\dfrac{dx}{x}. Integrating: …

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