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Miscellaneous Exercise 6(II) · Q133

Q.Solve: dydx=2y−x2y+x\dfrac{dy}{dx}=\dfrac{2y-x}{2y+x}

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dydx=2y−x2y+x\dfrac{dy}{dx}=\dfrac{2y-x}{2y+x} is homogeneous. Put y=vxy=vx: v+xdvdx=2v−12v+1v+x\dfrac{dv}{dx}=\dfrac{2v-1}{2v+1}, so xdvdx=2v−1−v(2v+1)2v+1=−2v2+v−12v+1=−2v2−v+12v+1x\dfrac{dv}{dx}=\dfrac{2v-1-v(2v+1)}{2v+1}=\dfrac{-2v^2+v-1}{2v+1}=-\dfrac{2v^2-v+1}{2v+1}, i.e. 2v+12v2−v+1dv=−dxx\dfrac{2v+1}{2v^2-v+1}dv=-\dfrac{dx}{x}. Splitting the numerator to match the derivative of the denominator (4v−14v-1) plus a remainder, integrating gives 12log⁡(2v2−v+1)+37tan⁡−1 ⁣(4v−17)=−log⁡x+c1\tfrac12\log(2v^2-v+1)+\dfrac{3}{\sqrt7}\tan^{-1}\!\left(\dfrac{4v-1}{\sqrt7}\right)=-\log x+c_1. Multiplying by 22 and substituting v=y/xv=y/x (usi …

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