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

The solution for y^{''}-y^'-12y=-5e^{3x} is y=c_{1}e^{4x}+c_{2}e^{-3x}+5/6 e^{3x}

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

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

y = c_{1} e^{4 x} + c_{2} e^{- 3 x} + \frac{5}{6} e^{3 x}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{4 x} + c_{2} e^{- 3 x} + \frac{5}{6} e^{3 x}

y = c_{1} e^{4 x} + c_{2} e^{- 3 x} + \frac{5}{6} e^{3 x}

y^{^{''}} + 5 y^{^{'}} + 4 = 0

\frac{d ^{2} y}{d x ^{2}} + y = 3 cos (2 x)

y^{^{''}} + 2 y^{^{'}} = 3 + sin (t)

y^{^{''}} - 2 y + 10 y = e^{x} cos (3 x)

y^{^{''}} - 8 y^{^{'}} + 7 = 0, y (0) = 1, y^{^{'}} (0) = 2

What is the solution for y^{''}-y^'-12y=-5e^{3x} ?
The solution for y^{''}-y^'-12y=-5e^{3x} is y=c_{1}e^{4x}+c_{2}e^{-3x}+5/6 e^{3x}