y^{''}+4y^'=sin(x)

Lösung

y^{^{''}} + 4 y^{^{'}} = sin (x)

y = c_{1} + c_{2} e^{- 4 x} - \frac{1}{17} sin (x) - \frac{4}{17} cos (x)

Schritte zur Lösung

y^{''} (x) + 4 y^{'} (x) = sin (x)

Solve linear ODE:

y = c_{1} + c_{2} e^{- 4 x} - \frac{1}{17} sin (x) - \frac{4}{17} cos (x)

y = c_{1} + c_{2} e^{- 4 x} - \frac{1}{17} sin (x) - \frac{4}{17} cos (x)

y^{^{''}} - 2 y^{^{'}} + y = \frac{t}{e ^{t}} + \frac{e ^{t}}{t}

4 y^{^{''}} - 7 y^{^{'}} + 3 y = t e^{2 t} sin (- 2 t)

(D^{2} - 3 D + 2) y = e^{5 x}

y^{^{''''}} - 2 y^{^{'''}} + y^{^{''}} = e^{x + 1}

y^{^{''}} + 3 y^{^{'}} + 6 = 0