y^'+y=xy^2

Lösung

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

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

Schritte zur Lösung

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

Solve Bernoulli ODE:

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

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

x \to \infty lim (2 x^{2} + 5)

\int \frac{x ^{2} - x + 2}{x ^{3} - x ^{2} + x - 1} d x

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

t an g e n t f (x) = \frac{1}{4 x ^{\frac{3}{2}}}, at x = 4

\frac{d y}{d x} = 6 x^{2} y