(20y^2-xy)dx+x^2dy=0

Lösung

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

y = \frac{x}{20 ln ( x ) + c _{1}}

Schritte zur Lösung

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

Solve Bernoulli ODE:

y = \frac{x}{20 ln ( x ) + c _{1}}

y = \frac{x}{20 ln ( x ) + c _{1}}

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