y^{''}+4y=2e^{3x}

Lösung

y^{^{''}} + 4 y = 2 e^{3 x}

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{2}{13} e^{3 x}

Schritte zur Lösung

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

Solve linear ODE:

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{2}{13} e^{3 x}

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{2}{13} e^{3 x}

y^{^{''}} + 25 y = 3 csc^{2} (5 t)

\frac{d ^{2} y}{d x ^{2}} - y = x

y^{^{''}} - 3 y^{^{'}} + 2 y = 12

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

(D^{2} - 4 D + 4) y = e^{x}