What is the solution for y^{''}+4y^'+5y=10 ?

The solution for y^{''}+4y^'+5y=10 is y=e^{-2t}(c_{1}cos(t)+c_{2}sin(t))+2

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y^{''}+4y^'+5y=10

y^{^{''}} + 4 y^{^{'}} + 5 y = 10

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

Solution steps

y^{''} (t) + 4 y^{'} (t) + 5 y = 10

Solve linear ODE:

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

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

What is the solution for y^{''}+4y^'+5y=10 ?
The solution for y^{''}+4y^'+5y=10 is y=e^{-2t}(c_{1}cos(t)+c_{2}sin(t))+2