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Exercise

dydx=2xy3−2xy\frac{dy}{dx}=2xy^3-2xy

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Final answer to the exercise

−ln⁡∣y∣+12ln⁡∣y+1∣+12ln⁡∣y−1∣=x2+C0-\ln\left|y\right|+\frac{1}{2}\ln\left|y+1\right|+\frac{1}{2}\ln\left|y-1\right|=x^2+C_0

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dydx =2xy3−2xy
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