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

Lösung

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

y = \frac{c _{1} + x ^{5} + x ^{4}}{2 x x} - \frac{x}{2}, y = - \frac{c _{1} + x ^{5} + x ^{4}}{2 x x} - \frac{x}{2}

Schritte zur Lösung

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

Solve using substitution:

y = \frac{c _{1} + x ^{5} + x ^{4}}{2 x x} - \frac{x}{2}, y = - \frac{c _{1} + x ^{5} + x ^{4}}{2 x x} - \frac{x}{2}

y = \frac{c _{1} + x ^{5} + x ^{4}}{2 x x} - \frac{x}{2}, y = - \frac{c _{1} + x ^{5} + x ^{4}}{2 x x} - \frac{x}{2}

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