What is the solution for 2x^{''}+5x^'+2x=4e-2t-t ?

The solution for 2x^{''}+5x^'+2x=4e-2t-t is x=c_{1}e^{-t/2}+c_{2}e^{-2t}-(3t)/2+(8e+15)/4

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2x^{''}+5x^'+2x=4e-2t-t

2 x^{^{''}} + 5 x^{^{'}} + 2 x = 4 e - 2 t - t

x = c_{1} e^{- \frac{t}{2}} + c_{2} e^{- 2 t} - \frac{3 t}{2} + \frac{8 e + 15}{4}

Solution steps

2 x^{''} (t) + 5 x^{'} (t) + 2 x = 4 e - 2 t - t

Solve linear ODE:

x = c_{1} e^{- \frac{t}{2}} + c_{2} e^{- 2 t} - \frac{3 t}{2} + \frac{8 e + 15}{4}

x = c_{1} e^{- \frac{t}{2}} + c_{2} e^{- 2 t} - \frac{3 t}{2} + \frac{8 e + 15}{4}

y^{^{''}} - 5 y^{^{'}} = 2 x + 1

y^{^{'''}} - 11 y^{^{''}} + 28 y^{^{'}} = 90 e^{x}

y^{^{''}} + y^{^{'}} = arctan (x)

y^{^{''}} - 2 y^{^{'}} - 8 y = 5 cos (t)

y^{^{''}} + 2 y^{^{'}} - 8 y = 8 x^{2} - 3

What is the solution for 2x^{''}+5x^'+2x=4e-2t-t ?
The solution for 2x^{''}+5x^'+2x=4e-2t-t is x=c_{1}e^{-t/2}+c_{2}e^{-2t}-(3t)/2+(8e+15)/4