What is the solution for y^{''}+8y^'+16=24x+2 ?

The solution for y^{''}+8y^'+16=24x+2 is y=c_{1}+c_{2}e^{-8x}+(3x^2)/2-(17x)/8

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y^{''}+8y^'+16=24x+2

y^{^{''}} + 8 y^{^{'}} + 16 = 24 x + 2

y = c_{1} + c_{2} e^{- 8 x} + \frac{3 x ^{2}}{2} - \frac{17 x}{8}

Solution steps

y^{''} (x) + 8 y^{'} (x) + 16 = 24 x + 2

Solve linear ODE:

y = c_{1} + c_{2} e^{- 8 x} + \frac{3 x ^{2}}{2} - \frac{17 x}{8}

y = c_{1} + c_{2} e^{- 8 x} + \frac{3 x ^{2}}{2} - \frac{17 x}{8}

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

y^{^{''}} - 4 y^{^{'}} + 4 y = x^{- 2} e^{2 x}

y^{^{''}} + 36 y = 6 tan (6 t) - 2 e^{3 t}

\frac{1}{2} y^{^{''}} + 10 y^{^{'}} + 100 y = 150

\frac{1}{6} y^{^{''}} + 6 y = 4 tan (6 t) - \frac{1}{2} e^{4 t}

What is the solution for y^{''}+8y^'+16=24x+2 ?
The solution for y^{''}+8y^'+16=24x+2 is y=c_{1}+c_{2}e^{-8x}+(3x^2)/2-(17x)/8