y^{''}+3y^'+2y=sin(t),y(0)=1,y^'(0)=1

Lösung

y^{^{''}} + 3 y^{^{'}} + 2 y = sin (t), y (0) = 1, y^{^{'}} (0) = 1

y = (\frac{11}{5} + \frac{13}{10}) e^{- t} - \frac{11}{5} e^{- 2 t} + \frac{1}{10} sin (t) - \frac{3}{10} cos (t)

Schritte zur Lösung

y^{''} (t) + 3 y^{'} (t) + 2 y = sin (t)

Solve linear ODE:

y = (\frac{11}{5} + \frac{13}{10}) e^{- t} - \frac{11}{5} e^{- 2 t} + \frac{1}{10} sin (t) - \frac{3}{10} cos (t)

y = (\frac{11}{5} + \frac{13}{10}) e^{- t} - \frac{11}{5} e^{- 2 t} + \frac{1}{10} sin (t) - \frac{3}{10} cos (t)

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

y^{^{''}} - 2 y^{^{'}} - 15 y = 3 x

(y^{^{''}} + 2 y^{^{'}} + 5) = e^{\frac{x}{3}}

y^{^{''}} + 5 y = - 180 x^{2} e^{(5 x)}

y^{^{''}} - 10 y^{^{'}} + 21 y = e^{7 t}