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

Lösung

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

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

Schritte zur Lösung

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

Solve Bernoulli ODE:

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

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

x \to 1 lim (\frac{x - 1}{x + 24 - 5})

\frac{\partial}{\partial x} (2 x + 4)

n = 1 \sum \infty \frac{n ^{4}}{n ^{5}}

\frac{\partial}{\partial z} (2 x - 3 yz)

\int \frac{5 x - 4}{x ^{2} + 6 x + 13} d x