What is the solution for y^{''}+14y^'+49y=4te^{-7t} ?

The solution for y^{''}+14y^'+49y=4te^{-7t} is y=c_{1}e^{-7t}+c_{2}te^{-7t}+(2e^{-7t}t^3)/3

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y^{''}+14y^'+49y=4te^{-7t}

y^{^{''}} + 14 y^{^{'}} + 49 y = 4 t e^{- 7 t}

y = c_{1} e^{- 7 t} + c_{2} t e^{- 7 t} + \frac{2 e ^{- 7 t} t ^{3}}{3}

Solution steps

y^{''} (t) + 14 y^{'} (t) + 49 y = 4 t e^{- 7 t}

Solve linear ODE:

y = c_{1} e^{- 7 t} + c_{2} t e^{- 7 t} + \frac{2 e ^{- 7 t} t ^{3}}{3}

y = c_{1} e^{- 7 t} + c_{2} t e^{- 7 t} + \frac{2 e ^{- 7 t} t ^{3}}{3}

4 y^{^{''}} - 4 y^{^{'}} + y = 16 e^{\frac{x}{2}}

y^{^{''}} - y^{^{'}} = e^{x} sin (x)

y^{^{''}} + 2 y^{^{'}} - 3 y = cos (2 x)

y^{^{'''}} - y^{^{''}} - 4 y^{^{'}} + 4 y = 20 e^{3 t}

y^{^{''}} - 8 y^{^{'}} + 17 y = 10 e^{4 x} sec (x)

What is the solution for y^{''}+14y^'+49y=4te^{-7t} ?
The solution for y^{''}+14y^'+49y=4te^{-7t} is y=c_{1}e^{-7t}+c_{2}te^{-7t}+(2e^{-7t}t^3)/3