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

The solution for y^{''}+y^'-2y=e^{4x} is y=c_{1}e^x+c_{2}e^{-2x}+1/18 e^{4x}

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

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

y = c_{1} e^{x} + c_{2} e^{- 2 x} + \frac{1}{18} e^{4 x}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{x} + c_{2} e^{- 2 x} + \frac{1}{18} e^{4 x}

y = c_{1} e^{x} + c_{2} e^{- 2 x} + \frac{1}{18} e^{4 x}

y^{^{''}} + 4 y^{^{'}} + 4 y = 3 e^{- t}

y^{^{''}} - 4 y + 8 y = 4 sin (x)

y^{^{''}} - 8 y^{^{'}} \cdot + 25 = 0

y^{^{''}} + 4 y^{^{'}} - 5 y = 75 t^{3}

y^{^{''}} - 7 y^{^{'}} + 12 y = 5 sin (x)

What is the solution for y^{''}+y^'-2y=e^{4x} ?
The solution for y^{''}+y^'-2y=e^{4x} is y=c_{1}e^x+c_{2}e^{-2x}+1/18 e^{4x}