(dy)/(dx)=(2y+1)^2xe^{-x}

Lösung

\frac{d y}{d x} = (2 y + 1)^{2} x e^{- x}

y = - \frac{e ^{x}}{4 ( - x - 1 + c _{1} e ^{x} )} + \frac{x}{2 ( - x + c _{1} e ^{x} - 1 )} - \frac{c _{1} e ^{x}}{2 ( - x + c _{1} e ^{x} - 1 )} + \frac{1}{2 ( - x + c _{1} e ^{x} - 1 )}

Schritte zur Lösung

\frac{d y}{d x} = (2 y + 1)^{2} x e^{- x}

Solve separable ODE:

y = - \frac{e ^{x}}{4 ( - x - 1 + c _{1} e ^{x} )} + \frac{x}{2 ( - x + c _{1} e ^{x} - 1 )} - \frac{c _{1} e ^{x}}{2 ( - x + c _{1} e ^{x} - 1 )} + \frac{1}{2 ( - x + c _{1} e ^{x} - 1 )}

y = - \frac{e ^{x}}{4 ( - x - 1 + c _{1} e ^{x} )} + \frac{x}{2 ( - x + c _{1} e ^{x} - 1 )} - \frac{c _{1} e ^{x}}{2 ( - x + c _{1} e ^{x} - 1 )} + \frac{1}{2 ( - x + c _{1} e ^{x} - 1 )}

\frac{d x}{d t} = - 300 \cdot \frac{x}{1 0 ^{8}}

y^{^{'}} + \frac{x}{1 + x ^{2}} \cdot y = \frac{2 x}{1 + x ^{2}}

\frac{d y}{d x} + 9 x y = 8 x, y (0) = 2

\frac{d y}{d x} + 9 x y = 8 x, y (0) = 5

x^{^{'}} = a x (1 - \frac{x}{k})