What is the solution for y^{''}+2y^'+y=t^5e^{-t} ?

The solution for y^{''}+2y^'+y=t^5e^{-t} is y=c_{1}e^{-t}+c_{2}te^{-t}+(e^{-t}t^7)/(42)

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y^{''}+2y^'+y=t^5e^{-t}

y^{^{''}} + 2 y^{^{'}} + y = t^{5} e^{- t}

y = c_{1} e^{- t} + c_{2} t e^{- t} + \frac{e ^{- t} t ^{7}}{42}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- t} + c_{2} t e^{- t} + \frac{e ^{- t} t ^{7}}{42}

y = c_{1} e^{- t} + c_{2} t e^{- t} + \frac{e ^{- t} t ^{7}}{42}

y^{^{''}} + 49 y = 17 sec^{2} (7 t)

y^{^{''}} + 4 y = (5 x^{2} + 4 x - 3) e^{x}

(\frac{1}{2}) y^{^{''}} + y^{^{'}} + 5 y = sin (3 t)

y^{^{''''}} + 25 y^{^{''}} = sin (5 t) + t e^{6 t} + 4

6 y^{^{''}} - 10 y^{^{'}} - 4 y = x + 2

What is the solution for y^{''}+2y^'+y=t^5e^{-t} ?
The solution for y^{''}+2y^'+y=t^5e^{-t} is y=c_{1}e^{-t}+c_{2}te^{-t}+(e^{-t}t^7)/(42)