3x^2y^2dx+(2x^3y+4y^3)dy=0

Lösung

3 x^{2} y^{2} d x + (2 x^{3} y + 4 y^{3}) d y = 0

y = \frac{- x ^{3} + x ^{6} + c _{1}}{2}, y = - \frac{- x ^{3} + x ^{6} + c _{1}}{2}, y = \frac{- x ^{3} - x ^{6} + c _{1}}{2}, y = - \frac{- x ^{3} - x ^{6} + c _{1}}{2}

Schritte zur Lösung

3 x^{2} y^{2} d x + (2 x^{3} y + 4 y^{3}) d y = 0

Solve exact ODE:

y = \frac{- x ^{3} + x ^{6} + c _{1}}{2}, y = - \frac{- x ^{3} + x ^{6} + c _{1}}{2}, y = \frac{- x ^{3} - x ^{6} + c _{1}}{2}, y = - \frac{- x ^{3} - x ^{6} + c _{1}}{2}

y = \frac{- x ^{3} + x ^{6} + c _{1}}{2}, y = - \frac{- x ^{3} + x ^{6} + c _{1}}{2}, y = \frac{- x ^{3} - x ^{6} + c _{1}}{2}, y = - \frac{- x ^{3} - x ^{6} + c _{1}}{2}

\int (x + 3) (x - 5) d x

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in v erse l a pl a ce \frac{1}{( s ^{2} + 4 ) \cdot s ^{2} \cdot e}

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