y^'+2y+3=0

Lösung

y^{^{'}} + 2 y + 3 = 0

y = \frac{e ^{- 2 t - c_{1}}}{2} - \frac{3}{2}

Schritte zur Lösung

y^{'} (t) + 2 y + 3 = 0

Solve separable ODE:

y = \frac{e ^{- 2 t - c_{1}}}{2} - \frac{3}{2}

y = \frac{e ^{- 2 t - c_{1}}}{2} - \frac{3}{2}

\frac{d v}{d t} = 0.4 - \frac{1}{30} v

y^{^{'}} + 9 x y = 8 x, y (0) = 4

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

x y^{^{'}} = y^{2} ln (x)

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