(x^2-1)y^'+2xy^2=0

Lösung

(x^{2} - 1) y^{^{'}} + 2 x y^{2} = 0

y = - \frac{1}{- ln ( x ^{2} - 1 ) + c _{1}}

Schritte zur Lösung

(x^{2} - 1) y^{'} (x) + 2 x y^{2} = 0

Solve separable ODE:

y = - \frac{1}{- ln ( x ^{2} - 1 ) + c _{1}}

y = - \frac{1}{- ln ( x ^{2} - 1 ) + c _{1}}

n = 1 \sum \infty \frac{4}{n ( n + 1 )}

\frac{d}{d x} (ln (x) + \frac{1}{ln ( x )})

\frac{\partial}{\partial x} (x e^{x^{2} y^{2}})

in v erse l a pl a ce \frac{1}{s ( s ^{2} + 6 s + 13 )}

x \to 1 lim (10 x + 15)