y^'=y(xy^2+1)

Lösung

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

y = \frac{2 e ^{2 x}}{- 2 x e ^{2 x} + e ^{2 x} + c _{1}}, y = - \frac{2 e ^{2 x}}{- 2 x e ^{2 x} + e ^{2 x} + c _{1}}

Schritte zur Lösung

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

Solve Bernoulli ODE:

y = \frac{2 e ^{2 x}}{- 2 x e ^{2 x} + e ^{2 x} + c _{1}}, y = - \frac{2 e ^{2 x}}{- 2 x e ^{2 x} + e ^{2 x} + c _{1}}

y = \frac{2 e ^{2 x}}{- 2 x e ^{2 x} + e ^{2 x} + c _{1}}, y = - \frac{2 e ^{2 x}}{- 2 x e ^{2 x} + e ^{2 x} + c _{1}}

\int 4 sin^{6} (x) cos^{3} (x) d x

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

t an g e n t f (x) = x^{3} - 6 x^{2} + 12 x - 9

\frac{\partial}{\partial y} (- 0.2 y + 23)

\frac{\partial}{\partial x} (- 5 x y (9 - x - y))