What is the solution for y^{''}+2y^'-8y=x+1 ?

The solution for y^{''}+2y^'-8y=x+1 is y=c_{1}e^{2x}+c_{2}e^{-4x}-x/8-5/32

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y^{''}+2y^'-8y=x+1

y^{^{''}} + 2 y^{^{'}} - 8 y = x + 1

y = c_{1} e^{2 x} + c_{2} e^{- 4 x} - \frac{x}{8} - \frac{5}{32}

Solution steps

y^{''} (x) + 2 y^{'} (x) - 8 y = x + 1

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{- 4 x} - \frac{x}{8} - \frac{5}{32}

y = c_{1} e^{2 x} + c_{2} e^{- 4 x} - \frac{x}{8} - \frac{5}{32}

y^{^{''}} - y^{^{'}} - 6 y = 15 sin (3 x)

y^{^{''}} - 4 y^{^{'}} + 3 y = 10 e^{3 x}

y^{^{''}} - y^{^{'}} = 1 + 2 x + e^{x} + sin (x) + cos (x)

3 y^{^{''}} - 2 y^{^{'}} - 8 y = (2 x + 3) e^{5 x}

\frac{d ^{2} y}{d x ^{2}} - y = \frac{1}{x ^{2}}

What is the solution for y^{''}+2y^'-8y=x+1 ?
The solution for y^{''}+2y^'-8y=x+1 is y=c_{1}e^{2x}+c_{2}e^{-4x}-x/8-5/32