y^'=(y^4x^4+4x^4)/(y^3)

Lösung

y^{^{'}} = \frac{y ^{4} x ^{4} + 4 x ^{4}}{y ^{3}}

y = 4 e^{c_{1} + \frac{4 x ^{5}}{5}} - 4, y = - 4 e^{c_{1} + \frac{4 x ^{5}}{5}} - 4

Schritte zur Lösung

y^{'} (x) = \frac{y ^{4} x ^{4} + 4 x ^{4}}{y ^{3}}

Solve separable ODE:

y = 4 e^{c_{1} + \frac{4 x ^{5}}{5}} - 4, y = - 4 e^{c_{1} + \frac{4 x ^{5}}{5}} - 4

y = 4 e^{c_{1} + \frac{4 x ^{5}}{5}} - 4, y = - 4 e^{c_{1} + \frac{4 x ^{5}}{5}} - 4

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