y^'=-8y+16

Lösung

y^{^{'}} = - 8 y + 16

y = e^{- 8 t - c_{1}} + 2

Schritte zur Lösung

y^{'} (t) = - 8 y + 16

Solve separable ODE:

y = e^{- 8 t - c_{1}} + 2

y = e^{- 8 t - c_{1}} + 2

\frac{d y}{d x} = 0.5 (y e^{x})^{\frac{1}{3}}

y^{^{'}} = 6 y \cdot ln (\frac{2}{y})

u^{^{'}} = 2 x H (x)

y^{^{'}} = \frac{7 cos ( 7 x )}{3 + 2 y}, y (0) = - 1

y^{^{'}} = y \cdot ln (y)