y^{''}-9y=4e^{3t},y(0)=0,y^'(0)=-1

Lösung

y^{^{''}} - 9 y = 4 e^{3 t}, y (0) = 0, y^{^{'}} (0) = - 1

y = - \frac{5}{18} e^{3 t} + \frac{5}{18} e^{- 3 t} + \frac{2 e ^{3 t} t}{3}

Schritte zur Lösung

y^{''} (t) - 9 y = 4 e^{3 t}

Solve linear ODE:

y = - \frac{5}{18} e^{3 t} + \frac{5}{18} e^{- 3 t} + \frac{2 e ^{3 t} t}{3}

y = - \frac{5}{18} e^{3 t} + \frac{5}{18} e^{- 3 t} + \frac{2 e ^{3 t} t}{3}

y^{^{''}} - 3 y^{^{'}} + 2 y = 2 cos^{2} (4 x)

y^{^{'''}} + 2 y^{^{''}} + y^{^{'}} = 2 - 24 e^{x} + 40 e^{5 x}

y^{^{''}} + 2 y = cos (2) t + sin (2) t

y^{^{''}} + 25 y = 7 sec (5 x)

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