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

The solution for y^{''}-y^'-12y=3x-2+5e^x is y=c_{1}e^{4x}+c_{2}e^{-3x}-x/4+3/16-5/12 e^x

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y^{''}-y^'-12y=3x-2+5e^x

y^{^{''}} - y^{^{'}} - 12 y = 3 x - 2 + 5 e^{x}

y = c_{1} e^{4 x} + c_{2} e^{- 3 x} - \frac{x}{4} + \frac{3}{16} - \frac{5}{12} e^{x}

Solution steps

y^{''} (x) - y^{'} (x) - 12 y = 3 x - 2 + 5 e^{x}

Solve linear ODE:

y = c_{1} e^{4 x} + c_{2} e^{- 3 x} - \frac{x}{4} + \frac{3}{16} - \frac{5}{12} e^{x}

y = c_{1} e^{4 x} + c_{2} e^{- 3 x} - \frac{x}{4} + \frac{3}{16} - \frac{5}{12} e^{x}

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

x^{^{''}} + 5 x^{^{'}} + 4 x = cos (π t)

y^{^{''}} - 2 y^{^{'}} + y = t^{- 8} e^{t}

y^{^{''}} - 8 y^{^{'}} + 20 y = t e^{t}, y (0) = 0, y^{^{'}} (0) = 0

y^{^{'''}} - 2 y^{^{'}} - 4 y = 25 sin (x)

What is the solution for y^{''}-y^'-12y=3x-2+5e^x ?
The solution for y^{''}-y^'-12y=3x-2+5e^x is y=c_{1}e^{4x}+c_{2}e^{-3x}-x/4+3/16-5/12 e^x