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

Lösung

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

y = (\frac{5}{4} + \frac{23}{20}) e^{- t} - \frac{5}{4} e^{- 2 t} + \frac{9}{20} sin (2 t) - \frac{3}{20} cos (2 t)

Schritte zur Lösung

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

Solve linear ODE:

y = (\frac{5}{4} + \frac{23}{20}) e^{- t} - \frac{5}{4} e^{- 2 t} + \frac{9}{20} sin (2 t) - \frac{3}{20} cos (2 t)

y = (\frac{5}{4} + \frac{23}{20}) e^{- t} - \frac{5}{4} e^{- 2 t} + \frac{9}{20} sin (2 t) - \frac{3}{20} cos (2 t)

y^{^{''}} + y = 3 cos (t)

y^{^{''}} - 16 y = sin (2 x) - 3 cos (2 x)

y^{^{''}} - 6 y^{^{'}} + 9 y = e^{3 x} lo g_{e} (1 + x^{2})

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

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