y^{''}+8y^'+2y=cos(x),y(0)=0,y^'(0)=1

Lösung

y^{^{''}} + 8 y^{^{'}} + 2 y = cos (x), y (0) = 0, y^{^{'}} (0) = 1

y = (\frac{53 + 14}{130 14} - \frac{1}{65}) e^{(- 4 + 14) x} - \frac{53 + 14}{130 14} e^{(- 4 - 14) x} + \frac{8}{65} sin (x) + \frac{1}{65} cos (x)

Schritte zur Lösung

y^{''} (x) + 8 y^{'} (x) + 2 y = cos (x)

Solve linear ODE:

y = (\frac{53 + 14}{130 14} - \frac{1}{65}) e^{(- 4 + 14) x} - \frac{53 + 14}{130 14} e^{(- 4 - 14) x} + \frac{8}{65} sin (x) + \frac{1}{65} cos (x)

y = (\frac{53 + 14}{130 14} - \frac{1}{65}) e^{(- 4 + 14) x} - \frac{53 + 14}{130 14} e^{(- 4 - 14) x} + \frac{8}{65} sin (x) + \frac{1}{65} cos (x)

y^{^{''}} - 2 y^{^{'}} + y = x

y^{^{''}} - 3 y^{^{'}} + 2 y = 4 x + 6

y^{^{''}} + y^{^{'}} - 6 y = x e^{2 x}

\frac{d ^{2} y}{d x ^{2}} + y = sin (x)

y^{^{''}} - 16 y = 48 t - 24 e^{- 4 t}, y (0) = 0, y^{^{'}} (0) = 24