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

Lösung

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

y = c_{1} e^{3 t} + c_{2} e^{- t} - \frac{4}{65} sin (2 t) - \frac{7}{65} cos (2 t)

Schritte zur Lösung

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

Solve linear ODE:

y = c_{1} e^{3 t} + c_{2} e^{- t} - \frac{4}{65} sin (2 t) - \frac{7}{65} cos (2 t)

y = c_{1} e^{3 t} + c_{2} e^{- t} - \frac{4}{65} sin (2 t) - \frac{7}{65} cos (2 t)

y^{^{''}} - 5 y^{^{'}} + 6 y = 5 e^{2 x}

y^{^{'''}} - 5 y^{^{''}} + 6 y^{^{'}} = 0 + 6 sin (x)

y^{^{''}} + y = sin (x), y^{^{'}} (\frac{π}{2}) = \frac{3}{2}, y (0) = 0

\frac{d ^{3} y}{d x ^{3}} + \frac{d y}{d x} = tan (x)

3 x^{^{''}} + 25 x^{^{'}} + 48 x = 5 cos (4 t)