What is the solution for 5y^{''}+4y^'+8y=4x-1 ?

The solution for 5y^{''}+4y^'+8y=4x-1 is y=e^{-(2x)/5}(c_{1}cos((6x)/5)+c_{2}sin((6x)/5))+x/2-3/8

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5y^{''}+4y^'+8y=4x-1

5 y^{^{''}} + 4 y^{^{'}} + 8 y = 4 x - 1

y = e^{- \frac{2 x}{5}} (c_{1} cos (\frac{6 x}{5}) + c_{2} sin (\frac{6 x}{5})) + \frac{x}{2} - \frac{3}{8}

Solution steps

5 y^{''} (x) + 4 y^{'} (x) + 8 y = 4 x - 1

Solve linear ODE:

y = e^{- \frac{2 x}{5}} (c_{1} cos (\frac{6 x}{5}) + c_{2} sin (\frac{6 x}{5})) + \frac{x}{2} - \frac{3}{8}

y = e^{- \frac{2 x}{5}} (c_{1} cos (\frac{6 x}{5}) + c_{2} sin (\frac{6 x}{5})) + \frac{x}{2} - \frac{3}{8}

4 y^{^{''}} + 20 y^{^{'}} + 169 y = e^{2 t}

(D^{3} + 8) y = x^{4} + 2 x + 1

y^{^{''}} - 6 y^{^{'}} + 13 y = 13

y^{^{''}} + 4 y = 16 t

y^{^{''}} - 8 y^{^{'}} + 16 y = e^{4 t} ln (t)

What is the solution for 5y^{''}+4y^'+8y=4x-1 ?
The solution for 5y^{''}+4y^'+8y=4x-1 is y=e^{-(2x)/5}(c_{1}cos((6x)/5)+c_{2}sin((6x)/5))+x/2-3/8