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

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

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y^{''}+8y^'=8x

y^{^{''}} + 8 y^{^{'}} = 8 x

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

Solution steps

y^{''} (x) + 8 y^{'} (x) = 8 x

Solve linear ODE:

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

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

y^{^{''}} - 2 y^{^{'}} = x^{3}

y^{^{''}} + 4 y = 2 sec^{3} (2 t)

y^{^{''}} - 2 y^{^{'}} - 3 y = 12 e^{x} - 12

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

\frac{d ^{2} y}{d t ^{2}} - 4 y = t cos (2 t)

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