What is the solution for (-8x^{''}-16x^'+24)=8e^{2t} ?

The solution for (-8x^{''}-16x^'+24)=8e^{2t} is x=c_{1}e^{-2t}+c_{2}-1/8 e^{2t}+(3t)/2

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(-8x^{''}-16x^'+24)=8e^{2t}

(- 8 x^{^{''}} - 16 x^{^{'}} + 24) = 8 e^{2 t}

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

Solution steps

(- 8 x^{''} (t) - 16 x^{'} (t) + 24) = 8 e^{2 t}

Solve linear ODE:

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

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

y^{^{''}} - y^{^{'}} + 2 y = 4 e^{x}

y^{^{''}} - 3 y^{^{'}} - 6 y = 2 e^{2 x} + 4

(4 d^{4} + 20 d^{3} + 35 d^{2} + 25 d + 6) y = o

2 y^{^{''}} + 4 y^{^{'}} + 8 y = 10 cos (2 t)

y^{^{''}} - 7 y^{^{'}} + 12 y = e^{x}

What is the solution for (-8x^{''}-16x^'+24)=8e^{2t} ?
The solution for (-8x^{''}-16x^'+24)=8e^{2t} is x=c_{1}e^{-2t}+c_{2}-1/8 e^{2t}+(3t)/2