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

The solution for y^{''}+5y^'=4x+8 is y=c_{1}+c_{2}e^{-5x}+(2x^2)/5+(36x)/(25)

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

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

y = c_{1} + c_{2} e^{- 5 x} + \frac{2 x ^{2}}{5} + \frac{36 x}{25}

Solution steps

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

Solve linear ODE:

y = c_{1} + c_{2} e^{- 5 x} + \frac{2 x ^{2}}{5} + \frac{36 x}{25}

y = c_{1} + c_{2} e^{- 5 x} + \frac{2 x ^{2}}{5} + \frac{36 x}{25}

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

y^{^{''}} + 4 y = 4 csc^{2} (2 t)

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

y^{^{'''}} - y^{^{''}} - 25 y^{^{'}} + 25 y = 7 - e^{x} + e^{5 x}

(D^{2} + D - 2) y = e^{x}

What is the solution for y^{''}+5y^'=4x+8 ?
The solution for y^{''}+5y^'=4x+8 is y=c_{1}+c_{2}e^{-5x}+(2x^2)/5+(36x)/(25)