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

The solution for y^{''}+5y^'=x+2 is y=c_{1}+c_{2}e^{-5x}+(x^2)/(10)+(9x)/(25)

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y^{''}+5y^'=x+2

y^{^{''}} + 5 y^{^{'}} = x + 2

y = c_{1} + c_{2} e^{- 5 x} + \frac{x ^{2}}{10} + \frac{9 x}{25}

Solution steps

y^{''} (x) + 5 y^{'} (x) = x + 2

Solve linear ODE:

y = c_{1} + c_{2} e^{- 5 x} + \frac{x ^{2}}{10} + \frac{9 x}{25}

y = c_{1} + c_{2} e^{- 5 x} + \frac{x ^{2}}{10} + \frac{9 x}{25}

y^{^{'''}} = 3 y^{^{''}} + 9 \cdot e^{x} + 0 \cdot cos (x)

y^{^{''}} + 4 y^{^{'}} + 4 y = e^{- 2 x l n (x)}

y^{^{'''}} - 7 y^{^{''}} + 14 y^{^{'}} - 8 y = ln (x)

y^{^{''}} + 5 y + 4 y = 5 e^{- 5 x}, y (0) = - 2, y^{^{'}} (0) = 2

y^{^{''}} + 5 y^{^{'}} + 4 y = 17 t e^{5 t}

What is the solution for y^{''}+5y^'=x+2 ?
The solution for y^{''}+5y^'=x+2 is y=c_{1}+c_{2}e^{-5x}+(x^2)/(10)+(9x)/(25)