x^'=(x^2+x)/(2x+1)

Lösung

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

x = \frac{- 1 + 1 + 4 e ^{t + c_{1}}}{2}, x = \frac{- 1 - 1 + 4 e ^{t + c_{1}}}{2}

Schritte zur Lösung

x^{'} (t) = \frac{x ^{2} + x}{2 x + 1}

Solve separable ODE:

x = \frac{- 1 + 1 + 4 e ^{t + c_{1}}}{2}, x = \frac{- 1 - 1 + 4 e ^{t + c_{1}}}{2}

x = \frac{- 1 + 1 + 4 e ^{t + c_{1}}}{2}, x = \frac{- 1 - 1 + 4 e ^{t + c_{1}}}{2}

(2 x^{3} - 1) d x - y d y = 0

(x - 1) d x - (x e^{y} + e^{y}) d y = 0

y^{^{'}} = k \frac{x}{y}

\frac{d y}{d t} - y = 0, y (0) = 1

\frac{d y}{d x} = \frac{y - 1}{x ^{2} 4 + x ^{2}}