5y^{''}+6y^'+25y=4cos(2t)

Lösung

5 y^{^{''}} + 6 y^{^{'}} + 25 y = 4 cos (2 t)

y = e^{- \frac{3 t}{5}} (c_{1} cos (\frac{2 29 t}{5}) + c_{2} sin (\frac{2 29 t}{5})) + \frac{48}{169} sin (2 t) + \frac{20}{169} cos (2 t)

Schritte zur Lösung

5 y^{''} (t) + 6 y^{'} (t) + 25 y = 4 cos (2 t)

Solve linear ODE:

y = e^{- \frac{3 t}{5}} (c_{1} cos (\frac{2 29 t}{5}) + c_{2} sin (\frac{2 29 t}{5})) + \frac{48}{169} sin (2 t) + \frac{20}{169} cos (2 t)

y = e^{- \frac{3 t}{5}} (c_{1} cos (\frac{2 29 t}{5}) + c_{2} sin (\frac{2 29 t}{5})) + \frac{48}{169} sin (2 t) + \frac{20}{169} cos (2 t)

y^{^{''}} + y^{^{'}} = 3, x^{2} y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} - 2 y^{^{'}} - 7 y = 1 e^{5 t}

9 y^{^{''}} + 4 y^{^{'}} - y = 22

x^{^{''}} + 4 x = 12 \cdot cos (t)

(D^{3} - 2 D^{2} - 5 D + 6) y = e^{4 x}