y^'+2xy=xy^2

Lösung

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

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

Schritte zur Lösung

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

Solve Bernoulli ODE:

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

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

\frac{\partial}{\partial x} (3 e^{x} sin (y))

t an g e n t f (x) = x - 2, at x = 9

l a pl a ce t r an s f or m 6 \cdot cos (0.2 t)

\frac{d y}{d x} = 3 y + 2

x \to 1 lim (sin (\frac{π}{2} x))