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

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

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y^{''}+5y^'+4y=8x

y^{^{''}} + 5 y^{^{'}} + 4 y = 8 x

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 5 y^{^{'}} + 6 y = e^{2 x} + 1

y^{^{''}} - y^{^{'}} - 2 y = 5 e^{4 x} + 6 x

\frac{d ^{2} y}{d x ^{2}} + \frac{d y}{d x} = x + e^{- 3 x}

y^{^{''}} + 4 y - 5 = 0

y^{^{''}} - 8 y^{^{'}} - 20 y = 270 e^{x}

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