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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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

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

y^{^{''}} + 2 y = 8 x + 5

y^{^{''}} + 9 y^{^{'}} = cos (3 t)

(D^{2} + 2 D + 1) y = x e^{- x}

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