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

The solution for y^{''}+y^'-6y=e^{4x} is y=c_{1}e^{2x}+c_{2}e^{-3x}+1/14 e^{4x}

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

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

y = c_{1} e^{2 x} + c_{2} e^{- 3 x} + \frac{1}{14} e^{4 x}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{- 3 x} + \frac{1}{14} e^{4 x}

y = c_{1} e^{2 x} + c_{2} e^{- 3 x} + \frac{1}{14} e^{4 x}

y^{^{''}} - 3 y^{^{'}} = 4 x

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

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

x^{^{''}} + 6 x^{^{'}} + 9 x = 10 cos (2 t)

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

What is the solution for y^{''}+y^'-6y=e^{4x} ?
The solution for y^{''}+y^'-6y=e^{4x} is y=c_{1}e^{2x}+c_{2}e^{-3x}+1/14 e^{4x}