y^{''}-2y^'-3y=et,y(0)=2,y^'(0)=4

Lösung

y^{^{''}} - 2 y^{^{'}} - 3 y = e t, y (0) = 2, y^{^{'}} (0) = 4

y = (2 + \frac{e - 2}{4} - \frac{2 e}{9}) e^{3 t} - \frac{e - 2}{4} e^{- t} - \frac{e t}{3} + \frac{2 e}{9}

Schritte zur Lösung

y^{''} (t) - 2 y^{'} (t) - 3 y = e t

Solve linear ODE:

y = (2 + \frac{e - 2}{4} - \frac{2 e}{9}) e^{3 t} - \frac{e - 2}{4} e^{- t} - \frac{e t}{3} + \frac{2 e}{9}

y = (2 + \frac{e - 2}{4} - \frac{2 e}{9}) e^{3 t} - \frac{e - 2}{4} e^{- t} - \frac{e t}{3} + \frac{2 e}{9}

x^{^{''}} + 23.04 x = 2 cos (5 t)

y^{^{''}} + 4 y = 9, t y (0) = 0, y^{^{'}} (0) = 7

y^{^{''}} + 8 y^{^{'}} + 12 y = 1 - H (t)

y^{^{'''}} - 11 y^{^{''}} + 28 y^{^{'}} = 36 e^{x}, y (0) = 25

y^{^{''}} - 2 y^{^{'}} + 5 y = e x sin (x)