What is the solution for (d^2x)/(dt^2)+6(dx)/(dt)+20x=5\H(t) ?

The solution for (d^2x)/(dt^2)+6(dx)/(dt)+20x=5\H(t) is

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(d^2x)/(dt^2)+6(dx)/(dt)+20x=5\H(t)

\frac{d ^{2} x}{d t ^{2}} + 6 \frac{d x}{d t} + 20 x = 5 H (t)

x = e^{- 3 t} (c_{1} cos (11 t) + c_{2} sin (11 t)) + e^{- 3 t} cos (11 t) \cdot \int - \frac{5 11 e ^{3 t} sin ( 11 t ) H ( t )}{11} d t + e^{- 3 t} sin (11 t) \cdot \int \frac{5 11 e ^{3 t} cos ( 11 t ) H ( t )}{11} d t

Solution steps

\frac{d ^{2} x}{d t ^{2}} + 6 \frac{d x}{d t} + 20 x = 5 H (t)

Solve linear ODE:

x = e^{- 3 t} (c_{1} cos (11 t) + c_{2} sin (11 t)) + e^{- 3 t} cos (11 t) \cdot \int - \frac{5 11 e ^{3 t} sin ( 11 t ) H ( t )}{11} d t + e^{- 3 t} sin (11 t) \cdot \int \frac{5 11 e ^{3 t} cos ( 11 t ) H ( t )}{11} d t

x = e^{- 3 t} (c_{1} cos (11 t) + c_{2} sin (11 t)) + e^{- 3 t} cos (11 t) \cdot \int - \frac{5 11 e ^{3 t} sin ( 11 t ) H ( t )}{11} d t + e^{- 3 t} sin (11 t) \cdot \int \frac{5 11 e ^{3 t} cos ( 11 t ) H ( t )}{11} d t

\frac{1}{2} y^{^{''}} + 10 y = 8 cos (6 t)

2 y^{^{''}} + 24 y^{^{'}} + \frac{1}{5} y = 12

0.1 q^{^{''}} + \frac{1}{1 0 ^{- 3}} q = 120 cos (101 t)

y^{^{''}} - 6 y^{^{'}} - 6 y = 9 x

(D^{3} + D) y = sec (x)

What is the solution for (d^2x)/(dt^2)+6(dx)/(dt)+20x=5\H(t) ?
The solution for (d^2x)/(dt^2)+6(dx)/(dt)+20x=5\H(t) is