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

Lösung

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

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

Schritte zur Lösung

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

Solve using substitution:

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

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

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