What is the solution for y^{''}+2y^'+5y=25 ?

The solution for y^{''}+2y^'+5y=25 is y=e^{-t}(c_{1}cos(2t)+c_{2}sin(2t))+5

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y^{''}+2y^'+5y=25

y^{^{''}} + 2 y^{^{'}} + 5 y = 25

y = e^{- t} (c_{1} cos (2 t) + c_{2} sin (2 t)) + 5

Solution steps

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

Solve linear ODE:

y = e^{- t} (c_{1} cos (2 t) + c_{2} sin (2 t)) + 5

y = e^{- t} (c_{1} cos (2 t) + c_{2} sin (2 t)) + 5

y^{^{'''}} + 4 y^{^{''}} + 8 y^{^{'}} = 11 + 5 e^{2 x}

y^{^{''}} - 9 y^{^{'}} = 81 x^{2} + 7, y (0) = - 1, y^{^{'}} (0) = 2

y^{^{''}} - y^{^{'}} - 6 y = 14 sin (3)

y^{^{''}} - 4 y + 3 y = 9 x^{2}

y^{^{''}} + 6 y^{^{'}} - 40 y = - 2 e^{3 x} + cos (2 x)

What is the solution for y^{''}+2y^'+5y=25 ?
The solution for y^{''}+2y^'+5y=25 is y=e^{-t}(c_{1}cos(2t)+c_{2}sin(2t))+5