What is the solution for x^{''}+6x^'-4x=8e^t ?

The solution for x^{''}+6x^'-4x=8e^t is x=c_{1}e^{(-3+sqrt(13))t}+c_{2}e^{(-3-sqrt(13))t}+8/3 e^t

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x^{''}+6x^'-4x=8e^t

x^{^{''}} + 6 x^{^{'}} - 4 x = 8 e^{t}

x = c_{1} e^{(- 3 + 13) t} + c_{2} e^{(- 3 - 13) t} + \frac{8}{3} e^{t}

Solution steps

x^{''} (t) + 6 x^{'} (t) - 4 x = 8 e^{t}

Solve linear ODE:

x = c_{1} e^{(- 3 + 13) t} + c_{2} e^{(- 3 - 13) t} + \frac{8}{3} e^{t}

x = c_{1} e^{(- 3 + 13) t} + c_{2} e^{(- 3 - 13) t} + \frac{8}{3} e^{t}

x^{^{''}} + 4 x^{^{'}} + 43 x = 9 cos (6 t)

y^{^{''}} + 9 y = 2 cos (3 x) - 3 sin (3 x)

y^{^{''}} - 4 y^{^{'}} + 3 = 0

y^{^{''}} + 3 y = 6 x

y^{^{''}} - 3 y^{^{'}} + 2 = 0

What is the solution for x^{''}+6x^'-4x=8e^t ?
The solution for x^{''}+6x^'-4x=8e^t is x=c_{1}e^{(-3+sqrt(13))t}+c_{2}e^{(-3-sqrt(13))t}+8/3 e^t