y^'=y-y^3

Lösung

y^{^{'}} = y - y^{3}

y = \frac{1}{- e ^{- 2 t - c_{1}} + 1}, y = - \frac{1}{- e ^{- 2 t - c_{1}} + 1}

Schritte zur Lösung

y^{'} (t) = y - y^{3}

Solve Bernoulli ODE:

y = \frac{1}{- e ^{- 2 t - c_{1}} + 1}, y = - \frac{1}{- e ^{- 2 t - c_{1}} + 1}

y = \frac{1}{- e ^{- 2 t - c_{1}} + 1}, y = - \frac{1}{- e ^{- 2 t - c_{1}} + 1}

l a pl a ce t r an s f or m sin (3 t) cos (5 t)

(2 x - 1) d x + (5 y + 8) d y = 0

\int_{0}^{\frac{1}{3}} \frac{3}{1 + 9 x ^{2}} d x

\int_{- 5}^{1} ∣ x + 3 ∣ d x

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