What is the solution for y^{''}+2y^'+5y=3\H(t) ?

The solution for y^{''}+2y^'+5y=3\H(t) is

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y^{''}+2y^'+5y=3\H(t)

y^{^{''}} + 2 y^{^{'}} + 5 y = 3 H (t)

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

Solution steps

y^{''} (t) + 2 y^{'} (t) + 5 y = 3 H (t)

Solve linear ODE:

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

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

y^{^{''}} + 9 y = 12 cos (3 t), y (0) = 0, y^{^{'}} (0) = 6

y^{^{''}} + 4 y^{^{'}} + 5 = 4 e^{- 2 t} tan (t)

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

y^{^{''}} - 2 y^{^{'}} = 3, y (0) = - 3 e^{- x}

y^{^{''}} + 9 y^{^{'}} + 18 y = x + e^{- x}

What is the solution for y^{''}+2y^'+5y=3\H(t) ?
The solution for y^{''}+2y^'+5y=3\H(t) is