4y^{''}+12y^'+9=0

Lösung

4 y^{^{''}} + 12 y^{^{'}} + 9 = 0

y = c_{1} + c_{2} e^{- 3 t} - \frac{3 t}{4}

Schritte zur Lösung

4 y^{''} (t) + 12 y^{'} (t) + 9 = 0

Solve linear ODE:

y = c_{1} + c_{2} e^{- 3 t} - \frac{3 t}{4}

y = c_{1} + c_{2} e^{- 3 t} - \frac{3 t}{4}

y^{^{''}} - 16 y = 7 e^{4 x}

y^{^{'''}} - y^{^{''}} = t

y^{^{''}} - 10 y^{^{'}} + 25 y = x cos (5 x)

y^{^{''}} + 9 y = 3 x, y (0) = 1, y^{^{'}} (0) = 3

y^{^{''''}} - y = e^{x}