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Exercise

dydx=y−xx−2y\frac{dy}{dx}=\frac{y-x}{x-2y}

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Final answer to the exercise

−1(2)3ln⁡∣1+2yx∣+1(2)3ln⁡∣−1+2yx∣−12ln⁡∣−x2+2y22x2∣=ln⁡∣x∣+C0\frac{-1}{\sqrt{\left(2\right)^{3}}}\ln\left|1+\frac{\sqrt{2}y}{x}\right|+\frac{1}{\sqrt{\left(2\right)^{3}}}\ln\left|-1+\frac{\sqrt{2}y}{x}\right|-\frac{1}{2}\ln\left|\frac{-x^2+2y^2}{2x^2}\right|=\ln\left|x\right|+C_0

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dydx =y−xx−2y 
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