What is the solution for y^{''}-y^'-2y=6x+5e^{4x} ?

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

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y^{''}-y^'-2y=6x+5e^{4x}

y^{^{''}} - y^{^{'}} - 2 y = 6 x + 5 e^{4 x}

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

Solution steps

y^{''} (x) - y^{'} (x) - 2 y = 6 x + 5 e^{4 x}

Solve linear ODE:

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

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

y^{^{''}} - 4 y^{^{'}} - 12 y = e^{- 2 x} + 5 x + 2

y^{^{''}} - 4 y^{^{'}} + 5 y = 24 cos (x)

y^{^{'''}} + 3 y^{^{''}} - 4 = 5 t^{2} + 4

y^{^{''}} - y^{^{'}} = - 11

y^{^{''}} - y^{^{'}} = - 10

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