What is the solution for y^{''}+y^'-6y=3x+2 ?

The solution for y^{''}+y^'-6y=3x+2 is y=c_{1}e^{2x}+c_{2}e^{-3x}-x/2-5/12

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y^{''}+y^'-6y=3x+2

y^{^{''}} + y^{^{'}} - 6 y = 3 x + 2

y = c_{1} e^{2 x} + c_{2} e^{- 3 x} - \frac{x}{2} - \frac{5}{12}

Solution steps

y^{''} (x) + y^{'} (x) - 6 y = 3 x + 2

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{- 3 x} - \frac{x}{2} - \frac{5}{12}

y = c_{1} e^{2 x} + c_{2} e^{- 3 x} - \frac{x}{2} - \frac{5}{12}

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

y^{^{''}} - 2 y^{^{'}} - 2 y = - 1 + 5 x^{2} + x^{3}

y^{^{''}} - 8 y^{^{'}} + 15 y = t e^{3 t}

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

y^{^{''}} + k^{2} y = a cos (b x), y (0) = 0, y^{^{'}} (0) = 0

What is the solution for y^{''}+y^'-6y=3x+2 ?
The solution for y^{''}+y^'-6y=3x+2 is y=c_{1}e^{2x}+c_{2}e^{-3x}-x/2-5/12