What is the solution for 2y^{''}-4y^'-6y=3e^{2x} ?

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

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2y^{''}-4y^'-6y=3e^{2x}

2 y^{^{''}} - 4 y^{^{'}} - 6 y = 3 e^{2 x}

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

Solution steps

2 y^{''} (x) - 4 y^{'} (x) - 6 y = 3 e^{2 x}

Solve linear ODE:

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

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

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

y^{^{''}} + 16 y = 7 sec (4 x)

y^{^{''}} + y^{^{'}} + 2 y = - 25 x e^{x}, y^{^{'}} (0) = 0

y^{^{''}} - 4 y^{^{'}} + 13 = 0

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

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