What is the solution for y^{''}-3y^'-40y=210e^{2t} ?

The solution for y^{''}-3y^'-40y=210e^{2t} is y=c_{1}e^{8t}+c_{2}e^{-5t}-5e^{2t}

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y^{''}-3y^'-40y=210e^{2t}

y^{^{''}} - 3 y^{^{'}} - 40 y = 210 e^{2 t}

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

Solution steps

y^{''} (t) - 3 y^{'} (t) - 40 y = 210 e^{2 t}

Solve linear ODE:

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

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

What is the solution for y^{''}-3y^'-40y=210e^{2t} ?
The solution for y^{''}-3y^'-40y=210e^{2t} is y=c_{1}e^{8t}+c_{2}e^{-5t}-5e^{2t}