What is the solution for y^{''}+25y^'=-24te^{-t} ?

The solution for y^{''}+25y^'=-24te^{-t} is y=c_{1}+c_{2}e^{-25t}+e^{-t}t+23/24 e^{-t}

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y^{''}+25y^'=-24te^{-t}

y^{^{''}} + 25 y^{^{'}} = - 24 t e^{- t}

y = c_{1} + c_{2} e^{- 25 t} + e^{- t} t + \frac{23}{24} e^{- t}

Solution steps

y^{''} (t) + 25 y^{'} (t) = - 24 t e^{- t}

Solve linear ODE:

y = c_{1} + c_{2} e^{- 25 t} + e^{- t} t + \frac{23}{24} e^{- t}

y = c_{1} + c_{2} e^{- 25 t} + e^{- t} t + \frac{23}{24} e^{- t}

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

2 y^{^{''}} - 2 y^{^{'}} - 4 y = 2 e^{3 t}

2 x^{^{''}} + 3 x^{^{'}} + 4 x = - 5 cos (t) + sin (t)

y^{^{''}} + 2 y^{^{'}} + 2 y = e^{- x} \cdot cos (x)

y^{^{''}} + 6 y^{^{'}} + 5 y = 30

What is the solution for y^{''}+25y^'=-24te^{-t} ?
The solution for y^{''}+25y^'=-24te^{-t} is y=c_{1}+c_{2}e^{-25t}+e^{-t}t+23/24 e^{-t}