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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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