y^{''}+2y^'-1=0

Lösung

y^{^{''}} + 2 y^{^{'}} - 1 = 0

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

Schritte zur Lösung

y^{''} (t) + 2 y^{'} (t) - 1 = 0

Solve linear ODE:

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

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

y^{^{''}} + 4 y^{^{'}} + 4 y = x^{2} \cdot e^{(- 2 x)} + 1

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

y^{^{''}} - y = sin (t), < y (0) = 0, y^{^{'}} (0) = 1

5 y^{^{''}} + y^{^{'}} = - 2 x, y (0) = 0, y^{^{'}} (0) = - 20

4 y^{^{''}} - 4 y^{^{'}} + y = e^{\frac{x}{2}}