What is the solution for y^{''}+4y^'=8x+1 ?

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

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y^{''}+4y^'=8x+1

y^{^{''}} + 4 y^{^{'}} = 8 x + 1

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

Solution steps

y^{''} (x) + 4 y^{'} (x) = 8 x + 1

Solve linear ODE:

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

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

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