What is the solution for (8x^{''}+4x^'-24)=48e^{2t} ?

The solution for (8x^{''}+4x^'-24)=48e^{2t} is x=c_{1}+c_{2}e^{-t/2}+6/5 e^{2t}+6t

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(8x^{''}+4x^'-24)=48e^{2t}

(8 x^{^{''}} + 4 x^{^{'}} - 24) = 48 e^{2 t}

x = c_{1} + c_{2} e^{- \frac{t}{2}} + \frac{6}{5} e^{2 t} + 6 t

Solution steps

(8 x^{''} (t) + 4 x^{'} (t) - 24) = 48 e^{2 t}

Solve linear ODE:

x = c_{1} + c_{2} e^{- \frac{t}{2}} + \frac{6}{5} e^{2 t} + 6 t

x = c_{1} + c_{2} e^{- \frac{t}{2}} + \frac{6}{5} e^{2 t} + 6 t

y^{^{''}} + 2 y^{^{'}} = 2 e^{- 2 t}

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

y^{^{''}} - 5 y^{^{'}} + 6 y = x^{2} - 4

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

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

What is the solution for (8x^{''}+4x^'-24)=48e^{2t} ?
The solution for (8x^{''}+4x^'-24)=48e^{2t} is x=c_{1}+c_{2}e^{-t/2}+6/5 e^{2t}+6t