What is the solution for y^{''}-3y^'-4y=e^{2x} ?

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

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y^{''}-3y^'-4y=e^{2x}

y^{^{''}} - 3 y^{^{'}} - 4 y = e^{2 x}

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

Solution steps

y^{''} (x) - 3 y^{'} (x) - 4 y = e^{2 x}

Solve linear ODE:

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

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

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

y^{^{''}} + 4 y^{^{'}} + 4 y = x^{2} + 3

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

9 y^{^{''}} + 4 y = 3 x + 12 e^{- x}, y (0) = 1, y^{^{'}} (0) = 0

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

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