What is the solution for x^{''}-5x^'+4x=8e^t ?

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

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x^{''}-5x^'+4x=8e^t

x^{^{''}} - 5 x^{^{'}} + 4 x = 8 e^{t}

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

Solution steps

x^{''} (t) - 5 x^{'} (t) + 4 x = 8 e^{t}

Solve linear ODE:

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

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

y^{^{''}} + 9 y = 10 e^{- t}

y^{^{''}} - 7 y^{^{'}} + 6 y = (x - 2) e^{x}

y^{^{''}} - 2 y^{^{'}} + 37 y = e^{x} cos (6 x)

y^{^{''}} + 3 y^{^{'}} - 4 y = cos (2 x)

y^{^{'''}} - 3 y^{^{''}} + 3 y^{^{'}} - y = e^{x}

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