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

Lösung

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

y = \frac{- 4 x ^{3} + x ^{2} + c _{1}}{2 x} - \frac{1}{2}, y = - \frac{- 4 x ^{3} + x ^{2} + c _{1}}{2 x} - \frac{1}{2}

Schritte zur Lösung

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

Solve using substitution:

y = \frac{- 4 x ^{3} + x ^{2} + c _{1}}{2 x} - \frac{1}{2}, y = - \frac{- 4 x ^{3} + x ^{2} + c _{1}}{2 x} - \frac{1}{2}

y = \frac{- 4 x ^{3} + x ^{2} + c _{1}}{2 x} - \frac{1}{2}, y = - \frac{- 4 x ^{3} + x ^{2} + c _{1}}{2 x} - \frac{1}{2}

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