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}

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

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

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

y^{^{''}} + 3 y^{^{'}} + 2 y = cos (x) - sin (2 x)

x^{^{''}} + 4 x^{^{'}} + 5 x = e^{- 2} t sin (t)

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