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

Lösung

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

y = (\frac{11}{5} + \frac{3}{10}) e^{- x} - \frac{11}{5} e^{- 2 x} + \frac{1}{10} sin (x) - \frac{3}{10} cos (x)

Schritte zur Lösung

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

Solve linear ODE:

y = (\frac{11}{5} + \frac{3}{10}) e^{- x} - \frac{11}{5} e^{- 2 x} + \frac{1}{10} sin (x) - \frac{3}{10} cos (x)

y = (\frac{11}{5} + \frac{3}{10}) e^{- x} - \frac{11}{5} e^{- 2 x} + \frac{1}{10} sin (x) - \frac{3}{10} cos (x)

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

y^{^{''}} - 6 y^{^{'}} + 8 y = 3 sin (2 x)

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

y^{^{''}} + 2 y = - 4 x + 10

y^{^{''''}} + 2 y^{^{''}} + y = 9 + cos (2 t)