What is the solution for y^{''}+5y^'+4y=8x^2+e^{-x} ?

The solution for y^{''}+5y^'+4y=8x^2+e^{-x} is y=c_{1}e^{-x}+c_{2}e^{-4x}+2x^2-5x+21/4+(e^{-x}x)/3

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y^{''}+5y^'+4y=8x^2+e^{-x}

y^{^{''}} + 5 y^{^{'}} + 4 y = 8 x^{2} + e^{- x}

y = c_{1} e^{- x} + c_{2} e^{- 4 x} + 2 x^{2} - 5 x + \frac{21}{4} + \frac{e ^{- x} x}{3}

Solution steps

y^{''} (x) + 5 y^{'} (x) + 4 y = 8 x^{2} + e^{- x}

Solve linear ODE:

y = c_{1} e^{- x} + c_{2} e^{- 4 x} + 2 x^{2} - 5 x + \frac{21}{4} + \frac{e ^{- x} x}{3}

y = c_{1} e^{- x} + c_{2} e^{- 4 x} + 2 x^{2} - 5 x + \frac{21}{4} + \frac{e ^{- x} x}{3}

What is the solution for y^{''}+5y^'+4y=8x^2+e^{-x} ?
The solution for y^{''}+5y^'+4y=8x^2+e^{-x} is y=c_{1}e^{-x}+c_{2}e^{-4x}+2x^2-5x+21/4+(e^{-x}x)/3