What is the solution for y^{''}+4y^'+5y=5 ?

The solution for y^{''}+4y^'+5y=5 is y=e^{-2t}(c_{1}cos(t)+c_{2}sin(t))+1

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y^{''}+4y^'+5y=5

y^{^{''}} + 4 y^{^{'}} + 5 y = 5

y = e^{- 2 t} (c_{1} cos (t) + c_{2} sin (t)) + 1

Solution steps

y^{''} (t) + 4 y^{'} (t) + 5 y = 5

Solve linear ODE:

y = e^{- 2 t} (c_{1} cos (t) + c_{2} sin (t)) + 1

y = e^{- 2 t} (c_{1} cos (t) + c_{2} sin (t)) + 1

4 y^{^{''}} + 4 y^{^{'}} + y = \frac{5 e ^{- \frac{t}{2}}}{t ^{2} + 1}

y^{^{''}} - 5 y^{^{'}} + 6 y = - 3 sin (2 x)

y^{^{''}} + 4 y^{^{'}} + 3 y = 18

y^{^{''}} + y^{^{'}} + y = - 13 sin (2 t)

y^{^{''}} - 7 y^{^{'}} + 12 y = 2 e^{4 t}, y (0) = 1, y^{^{'}} (0) = 0

What is the solution for y^{''}+4y^'+5y=5 ?
The solution for y^{''}+4y^'+5y=5 is y=e^{-2t}(c_{1}cos(t)+c_{2}sin(t))+1