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

Lösung

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

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

Schritte zur Lösung

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

Solve using substitution:

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

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

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