What is the solution for y^{''}-y^'-6y=e^{3x}+30 ?

The solution for y^{''}-y^'-6y=e^{3x}+30 is y=c_{1}e^{3x}+c_{2}e^{-2x}+(e^{3x}x)/5-5

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y^{''}-y^'-6y=e^{3x}+30

y^{^{''}} - y^{^{'}} - 6 y = e^{3 x} + 30

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

Solution steps

y^{''} (x) - y^{'} (x) - 6 y = e^{3 x} + 30

Solve linear ODE:

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

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

y^{^{''}} + 4 y = 1

y^{^{'''}} + y^{^{''}} = e^{x}

y^{^{''}} + 3 y^{^{'}} + 2 y = \frac{1}{1 + e ^{2 x}}

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

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

What is the solution for y^{''}-y^'-6y=e^{3x}+30 ?
The solution for y^{''}-y^'-6y=e^{3x}+30 is y=c_{1}e^{3x}+c_{2}e^{-2x}+(e^{3x}x)/5-5