What is the solution for y^{''}-3y^'-40y=168e^{2t} ?

The solution for y^{''}-3y^'-40y=168e^{2t} is y=c_{1}e^{8t}+c_{2}e^{-5t}-4e^{2t}

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y^{''}-3y^'-40y=168e^{2t}

y^{^{''}} - 3 y^{^{'}} - 40 y = 168 e^{2 t}

y = c_{1} e^{8 t} + c_{2} e^{- 5 t} - 4 e^{2 t}

Solution steps

y^{''} (t) - 3 y^{'} (t) - 40 y = 168 e^{2 t}

Solve linear ODE:

y = c_{1} e^{8 t} + c_{2} e^{- 5 t} - 4 e^{2 t}

y = c_{1} e^{8 t} + c_{2} e^{- 5 t} - 4 e^{2 t}

y^{^{'''}} - y^{^{''}} - 4 y^{^{'}} + 4 y = sin (x)

y^{^{''}} + b y = a

y^{^{''}} + y^{^{'}} + x = 0

y^{^{''}} + 2 y^{^{'}} + 2 y = e^{- x} cot (x) sec (x)

2 y^{^{''}} - 7 y^{^{'}} + 5 y = 11 - 7 x - 16 x^{2} + 5 x^{3}

What is the solution for y^{''}-3y^'-40y=168e^{2t} ?
The solution for y^{''}-3y^'-40y=168e^{2t} is y=c_{1}e^{8t}+c_{2}e^{-5t}-4e^{2t}