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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 2 y^{^{'}} + y = 9 - e^{2 x}, y (0) = 0, y^{^{'}} (0) = 1

(3 d^{2} + 3) y = 3 x

y^{^{''}} + 5 y^{^{'}} + 6 y = 4 + 5 x

y^{^{''}} + 3 y^{^{'}} - 10 y = 20 + 5 e^{3 x}

x^{^{''}} + 4 x^{^{'}} + 14 x = 9 cos (3 t)

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