x^'=-ax+b

Lösung

x^{^{'}} = - a x + b

x = \frac{e ^{- t a - c_{1} a}}{a} + \frac{b}{a}

Schritte zur Lösung

x^{'} (t) = - a x + b

Solve separable ODE:

x = \frac{e ^{- t a - c_{1} a}}{a} + \frac{b}{a}

x = \frac{e ^{- t a - c_{1} a}}{a} + \frac{b}{a}

y e^{x} d x + e^{x} d y = 0

y^{^{'}} y^{2} x = y^{3}

2 + y^{^{'}} = 6 y, y (- 2) = 8

y^{^{'}} y = (4 x^{3} + 5) (4 x^{7} - 7)

4 (5 x - 3)^{2} \frac{d y}{d x} = 2 y - 4