y^'+2y-5=0

Lösung

y^{^{'}} + 2 y - 5 = 0

y = \frac{e ^{- 2 t - c_{1}}}{2} + \frac{5}{2}

Schritte zur Lösung

y^{'} (t) + 2 y - 5 = 0

Solve separable ODE:

y = \frac{e ^{- 2 t - c_{1}}}{2} + \frac{5}{2}

y = \frac{e ^{- 2 t - c_{1}}}{2} + \frac{5}{2}

\frac{d y}{d x} y = \frac{6 x ^{2} + 4 x ^{2}}{3 y ^{6}}

y^{2} y^{^{'}} = x^{2} + 6 e^{x}

y^{^{'}} + y = 0, y (1) = 1

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

y^{2} d y = x d x