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}

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