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

The solution for y^{''}-3y^'-4y=x is y=c_{1}e^{4x}+c_{2}e^{-x}-x/4+3/16

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

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

y = c_{1} e^{4 x} + c_{2} e^{- x} - \frac{x}{4} + \frac{3}{16}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{4 x} + c_{2} e^{- x} - \frac{x}{4} + \frac{3}{16}

y = c_{1} e^{4 x} + c_{2} e^{- x} - \frac{x}{4} + \frac{3}{16}

\frac{d ^{2} y}{d x ^{2}} - 10 \frac{d y}{d x} + 25 y = 30 x + 3

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

y^{^{''''}} + 18 y^{^{''}} + 81 y = 162 + cos (6 t)

\frac{d ^{2} y}{d x ^{2}} - 5 \frac{d y}{d x} + 6 y = 3 x

y^{^{''}} - y^{^{'}} - 12 y = 2 t^{6} e^{- 3 t}

What is the solution for y^{''}-3y^'-4y=x ?
The solution for y^{''}-3y^'-4y=x is y=c_{1}e^{4x}+c_{2}e^{-x}-x/4+3/16