What is the solution for y^{''}-y^'+1/4 y=3+e^{-2x} ?

The solution for y^{''}-y^'+1/4 y=3+e^{-2x} is y=c_{1}e^{x/2}+c_{2}e^{x/2}x+12+4/25 e^{-2x}

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y^{''}-y^'+1/4 y=3+e^{-2x}

y^{^{''}} - y^{^{'}} + \frac{1}{4} y = 3 + e^{- 2 x}

y = c_{1} e^{\frac{x}{2}} + c_{2} e^{\frac{x}{2}} x + 12 + \frac{4}{25} e^{- 2 x}

Solution steps

y^{''} (x) - y^{'} (x) + \frac{1}{4} y = 3 + e^{- 2 x}

Solve linear ODE:

y = c_{1} e^{\frac{x}{2}} + c_{2} e^{\frac{x}{2}} x + 12 + \frac{4}{25} e^{- 2 x}

y = c_{1} e^{\frac{x}{2}} + c_{2} e^{\frac{x}{2}} x + 12 + \frac{4}{25} e^{- 2 x}

What is the solution for y^{''}-y^'+1/4 y=3+e^{-2x} ?
The solution for y^{''}-y^'+1/4 y=3+e^{-2x} is y=c_{1}e^{x/2}+c_{2}e^{x/2}x+12+4/25 e^{-2x}