y^'+y=y^2

Lösung

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

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

Schritte zur Lösung

y^{'} (t) + y = y^{2}

Solve separable ODE:

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

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

\frac{d}{d x} (30 x^{4} y^{7} - 24 x^{12} y^{3})

\frac{d}{d x} (- e^{x^{2}})

\frac{\partial}{\partial y} (x + ln (y))

\int_{1}^{e^{5}} x^{4} ln (x) d x

d er i v a t i v e y = \frac{x ^{2} - 1}{x ^{2} + 1}