y^{''}+2y^'+10y=e^{2x}cos(3x)

Lösung

y^{^{''}} + 2 y^{^{'}} + 10 y = e^{2 x} cos (3 x)

y = e^{- x} (c_{1} cos (3 x) + c_{2} sin (3 x)) + \frac{2}{45} e^{2 x} sin (3 x) + \frac{1}{45} e^{2 x} cos (3 x)

Schritte zur Lösung

y^{''} (x) + 2 y^{'} (x) + 10 y = e^{2 x} cos (3 x)

Solve linear ODE:

y = e^{- x} (c_{1} cos (3 x) + c_{2} sin (3 x)) + \frac{2}{45} e^{2 x} sin (3 x) + \frac{1}{45} e^{2 x} cos (3 x)

y = e^{- x} (c_{1} cos (3 x) + c_{2} sin (3 x)) + \frac{2}{45} e^{2 x} sin (3 x) + \frac{1}{45} e^{2 x} cos (3 x)

y^{^{''}} - 2 y^{^{'}} + 5 y = - 20 e^{x} sin (2 x)

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

y^{^{''''}} - y^{^{'}} = x^{3} + x^{2} + 1

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

3 x^{^{''}} + 48 x = 60 e^{- 4 t} cos (4 t)