What is the solution for y^{''}+5y^'+6y=4x+5 ?

The solution for y^{''}+5y^'+6y=4x+5 is y=c_{1}e^{-2x}+c_{2}e^{-3x}+(2x)/3+5/18

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y^{''}+5y^'+6y=4x+5

y^{^{''}} + 5 y^{^{'}} + 6 y = 4 x + 5

y = c_{1} e^{- 2 x} + c_{2} e^{- 3 x} + \frac{2 x}{3} + \frac{5}{18}

Solution steps

y^{''} (x) + 5 y^{'} (x) + 6 y = 4 x + 5

Solve linear ODE:

y = c_{1} e^{- 2 x} + c_{2} e^{- 3 x} + \frac{2 x}{3} + \frac{5}{18}

y = c_{1} e^{- 2 x} + c_{2} e^{- 3 x} + \frac{2 x}{3} + \frac{5}{18}

3 x^{^{''}} + 30 x^{^{'}} + 125 x = 350

y^{^{''}} + y = t^{2} + 1

y^{^{''}} + 4 y^{^{'}} + 4 y = 5 e^{3 t}

y^{^{''}} + y^{^{'}} + \frac{5}{4} y = 3 cos (x)

y^{^{''}} + y^{^{'}} - 6 y = 8 + 4 e^{t}

What is the solution for y^{''}+5y^'+6y=4x+5 ?
The solution for y^{''}+5y^'+6y=4x+5 is y=c_{1}e^{-2x}+c_{2}e^{-3x}+(2x)/3+5/18