2y^{''}+6y^'+10y=4cos(t)

Lösung

2 y^{^{''}} + 6 y^{^{'}} + 10 y = 4 cos (t)

y = e^{- \frac{3 t}{2}} (c_{1} cos (\frac{11 t}{2}) + c_{2} sin (\frac{11 t}{2})) + \frac{6}{25} sin (t) + \frac{8}{25} cos (t)

Schritte zur Lösung

2 y^{''} (t) + 6 y^{'} (t) + 10 y = 4 cos (t)

Solve linear ODE:

y = e^{- \frac{3 t}{2}} (c_{1} cos (\frac{11 t}{2}) + c_{2} sin (\frac{11 t}{2})) + \frac{6}{25} sin (t) + \frac{8}{25} cos (t)

y = e^{- \frac{3 t}{2}} (c_{1} cos (\frac{11 t}{2}) + c_{2} sin (\frac{11 t}{2})) + \frac{6}{25} sin (t) + \frac{8}{25} cos (t)

y^{^{''}} - 14 y^{^{'}} + 49 y = 2 x e^{7 x}

y^{^{''}} - 6 y^{^{'}} + 9 y = e^{3} x, y (0) = - 1, y^{^{'}} (0) = 1

\frac{d ^{2} y}{d t ^{2}} - 5 \frac{d y}{d t} + 6 = 0

- 5 y^{^{''}} - 2 y^{^{'}} + 24 = 0

l in e a r \frac{1}{4} y^{^{''}} + 4 y = tan (4 t) - \frac{1}{3} e^{3 t}