y=(2y^4+2x)y^'

Lösung

y = (2 y^{4} + 2 x) y^{^{'}}

y = - \frac{c _{1} + c _{1}^{2} + 4 x}{2}, y = - - \frac{c _{1} + c _{1}^{2} + 4 x}{2}, y = - \frac{c _{1} - c _{1}^{2} + 4 x}{2}, y = - - \frac{c _{1} - c _{1}^{2} + 4 x}{2}

Schritte zur Lösung

y = (2 y^{4} + 2 x) y^{'} (x)

Solve exact ODE:

y = - \frac{c _{1} + c _{1}^{2} + 4 x}{2}, y = - - \frac{c _{1} + c _{1}^{2} + 4 x}{2}, y = - \frac{c _{1} - c _{1}^{2} + 4 x}{2}, y = - - \frac{c _{1} - c _{1}^{2} + 4 x}{2}

y = - \frac{c _{1} + c _{1}^{2} + 4 x}{2}, y = - - \frac{c _{1} + c _{1}^{2} + 4 x}{2}, y = - \frac{c _{1} - c _{1}^{2} + 4 x}{2}, y = - - \frac{c _{1} - c _{1}^{2} + 4 x}{2}

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