What is the solution for y^{''}+2y^'+y=t ?

The solution for y^{''}+2y^'+y=t is y=c_{1}e^{-t}+c_{2}te^{-t}+t-2

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y^{''}+2y^'+y=t

y^{^{''}} + 2 y^{^{'}} + y = t

y = c_{1} e^{- t} + c_{2} t e^{- t} + t - 2

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- t} + c_{2} t e^{- t} + t - 2

y = c_{1} e^{- t} + c_{2} t e^{- t} + t - 2

y^{^{''}} + y = cos (x) + sec (x)

y^{^{''}} - 5 y^{^{'}} + 6 y = 2 e^{- x}

y^{^{''}} - 3 y^{^{'}} + 5 y = x e^{x}

y^{^{''}} - 2 y^{^{'}} - 24 y = - 64 t + 48 t^{2}

y^{^{''}} - 2 y^{^{'}} + y = 13 e^{x}

What is the solution for y^{''}+2y^'+y=t ?
The solution for y^{''}+2y^'+y=t is y=c_{1}e^{-t}+c_{2}te^{-t}+t-2