What is the solution for 5y^{''}-19y^'-4y=1+4x ?

The solution for 5y^{''}-19y^'-4y=1+4x is y=c_{1}e^{4x}+c_{2}e^{-x/5}-x+9/2

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5y^{''}-19y^'-4y=1+4x

5 y^{^{''}} - 19 y^{^{'}} - 4 y = 1 + 4 x

y = c_{1} e^{4 x} + c_{2} e^{- \frac{x}{5}} - x + \frac{9}{2}

Solution steps

5 y^{''} (x) - 19 y^{'} (x) - 4 y = 1 + 4 x

Solve linear ODE:

y = c_{1} e^{4 x} + c_{2} e^{- \frac{x}{5}} - x + \frac{9}{2}

y = c_{1} e^{4 x} + c_{2} e^{- \frac{x}{5}} - x + \frac{9}{2}

(d^{3} - 7 d - 6) y = 2 e^{x} cos (\frac{x}{2})

\frac{1}{4} y^{^{''}} + 4 y = 4 tan (4 t) - \frac{1}{4} e^{3 t}

y^{^{''}} + 5 y^{^{'}} + 6 y = 3 e^{- 2 x}

y^{^{''}} - 4 y - 12 y = 12 t

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

What is the solution for 5y^{''}-19y^'-4y=1+4x ?
The solution for 5y^{''}-19y^'-4y=1+4x is y=c_{1}e^{4x}+c_{2}e^{-x/5}-x+9/2