x^2y^'=y^2+xy

Lösung

x^{2} y^{^{'}} = y^{2} + x y

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

Schritte zur Lösung

x^{2} y^{'} (x) = y^{2} + x y

Solve Bernoulli ODE:

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

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

s l o p e (3, 3), (- 3, - 1)

x \to 2 - lim (2 - x)

\frac{d}{d x} ((5 x^{3} - 7)^{8})

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