What is the solution for y^{''}+2y^'+y=7e^{-t} ?

The solution for y^{''}+2y^'+y=7e^{-t} is y=c_{1}e^{-t}+c_{2}te^{-t}+(7e^{-t}t^2)/2

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y^{''}+2y^'+y=7e^{-t}

y^{^{''}} + 2 y^{^{'}} + y = 7 e^{- t}

y = c_{1} e^{- t} + c_{2} t e^{- t} + \frac{7 e ^{- t} t ^{2}}{2}

Solution steps

y^{''} (t) + 2 y^{'} (t) + y = 7 e^{- t}

Solve linear ODE:

y = c_{1} e^{- t} + c_{2} t e^{- t} + \frac{7 e ^{- t} t ^{2}}{2}

y = c_{1} e^{- t} + c_{2} t e^{- t} + \frac{7 e ^{- t} t ^{2}}{2}

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

\frac{d ^{2} y}{d t ^{2}} + 5 \frac{d y}{d t} + 4 y = 2 t

- y^{^{''}} + 2 y^{^{'}} + 3 y = 1, y (0) = 0, y (1) = 0

y^{^{''}} + 4 y^{^{'}} + 8 = 0

y^{^{''}} - 4 y^{^{'}} - 12 y = 10 e^{- 2 x}

What is the solution for y^{''}+2y^'+y=7e^{-t} ?
The solution for y^{''}+2y^'+y=7e^{-t} is y=c_{1}e^{-t}+c_{2}te^{-t}+(7e^{-t}t^2)/2