y^'=(y^2-1)x

Lösung

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

y = - \frac{e ^{c_{1}}}{e ^{x^{2}} - e ^{c_{1}}} - \frac{e ^{x^{2}}}{e ^{x^{2}} - e ^{c_{1}}}

Schritte zur Lösung

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

Solve Bernoulli ODE:

y = - \frac{e ^{c_{1}}}{e ^{x^{2}} - e ^{c_{1}}} - \frac{e ^{x^{2}}}{e ^{x^{2}} - e ^{c_{1}}}

y = - \frac{e ^{c_{1}}}{e ^{x^{2}} - e ^{c_{1}}} - \frac{e ^{x^{2}}}{e ^{x^{2}} - e ^{c_{1}}}

\frac{d}{d x} (\frac{x ^{2} + 3}{x ^{2} - 9 x})

\frac{\partial}{\partial u} (u \cdot sin (v))

\int_{- 2}^{1} \frac{1}{2 - x} d x

\frac{d}{d x} (x e^{- 6 x})

k = 2 \sum \infty \frac{3}{2 ^{k}}