What is the solution for y^{''}+4y^'+6y=1^{-t} ?

The solution for y^{''}+4y^'+6y=1^{-t} is y=e^{-2t}(c_{1}cos(sqrt(2)t)+c_{2}sin(sqrt(2)t))+1/6

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y^{''}+4y^'+6y=1^{-t}

y^{^{''}} + 4 y^{^{'}} + 6 y = 1^{- t}

y = e^{- 2 t} (c_{1} cos (2 t) + c_{2} sin (2 t)) + \frac{1}{6}

Solution steps

y^{''} (t) + 4 y^{'} (t) + 6 y = 1^{- t}

Solve linear ODE:

y = e^{- 2 t} (c_{1} cos (2 t) + c_{2} sin (2 t)) + \frac{1}{6}

y = e^{- 2 t} (c_{1} cos (2 t) + c_{2} sin (2 t)) + \frac{1}{6}

y^{^{''}} + y = t - \frac{π}{2}, y (\frac{π}{2}) = 0, y^{^{'}} (\frac{π}{2}) = 1

y^{^{''}} - y^{^{'}} - 2 y = (x + 1) e^{- x}

y^{^{''}} - 2 y^{^{'}} - 3 y = 2

y^{^{''}} + 6 y^{^{'}} + 9 y = \frac{e ^{- 3 x}}{( x ^{2} + 1 )}

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

What is the solution for y^{''}+4y^'+6y=1^{-t} ?
The solution for y^{''}+4y^'+6y=1^{-t} is y=e^{-2t}(c_{1}cos(sqrt(2)t)+c_{2}sin(sqrt(2)t))+1/6