What is the solution for y^{''}-2y^'-8y=6+3x ?

The solution for y^{''}-2y^'-8y=6+3x is y=c_{1}e^{4x}+c_{2}e^{-2x}-(3x)/8-21/32

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y^{''}-2y^'-8y=6+3x

y^{^{''}} - 2 y^{^{'}} - 8 y = 6 + 3 x

y = c_{1} e^{4 x} + c_{2} e^{- 2 x} - \frac{3 x}{8} - \frac{21}{32}

Solution steps

y^{''} (x) - 2 y^{'} (x) - 8 y = 6 + 3 x

Solve linear ODE:

y = c_{1} e^{4 x} + c_{2} e^{- 2 x} - \frac{3 x}{8} - \frac{21}{32}

y = c_{1} e^{4 x} + c_{2} e^{- 2 x} - \frac{3 x}{8} - \frac{21}{32}

What is the solution for y^{''}-2y^'-8y=6+3x ?
The solution for y^{''}-2y^'-8y=6+3x is y=c_{1}e^{4x}+c_{2}e^{-2x}-(3x)/8-21/32