What is the solution for y^{''}-3y^'-4y=4x ?

The solution for y^{''}-3y^'-4y=4x is y=c_{1}e^{4x}+c_{2}e^{-x}-x+3/4

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

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

y = c_{1} e^{4 x} + c_{2} e^{- x} - x + \frac{3}{4}

Solution steps

y^{''} (x) - 3 y^{'} (x) - 4 y = 4 x

Solve linear ODE:

y = c_{1} e^{4 x} + c_{2} e^{- x} - x + \frac{3}{4}

y = c_{1} e^{4 x} + c_{2} e^{- x} - x + \frac{3}{4}

y^{^{''}} - 2 y^{^{'}} + 5 y = - 8, y (0) = 2, y^{^{'}} (0) = 2

\frac{1}{2} y^{^{''}} + 2 y = 2 tan (2 t) - \frac{1}{2} e^{t}

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

y^{^{''}} - y = x^{2} e^{x} + 9

y^{^{''}} - 2 y^{^{'}} - 3 y = 24 t e^{2 t}

What is the solution for y^{''}-3y^'-4y=4x ?
The solution for y^{''}-3y^'-4y=4x is y=c_{1}e^{4x}+c_{2}e^{-x}-x+3/4