What is the solution for 5x^{''}+20x^'+20x=28 ?

The solution for 5x^{''}+20x^'+20x=28 is x=c_{1}e^{-2t}+c_{2}te^{-2t}+7/5

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5x^{''}+20x^'+20x=28

5 x^{^{''}} + 20 x^{^{'}} + 20 x = 28

x = c_{1} e^{- 2 t} + c_{2} t e^{- 2 t} + \frac{7}{5}

Solution steps

5 x^{''} (t) + 20 x^{'} (t) + 20 x = 28

Solve linear ODE:

x = c_{1} e^{- 2 t} + c_{2} t e^{- 2 t} + \frac{7}{5}

x = c_{1} e^{- 2 t} + c_{2} t e^{- 2 t} + \frac{7}{5}

What is the solution for 5x^{''}+20x^'+20x=28 ?
The solution for 5x^{''}+20x^'+20x=28 is x=c_{1}e^{-2t}+c_{2}te^{-2t}+7/5