What is the solution for y^{''}-6y^'+8y=-2e^{3x} ?

The solution for y^{''}-6y^'+8y=-2e^{3x} is y=c_{1}e^{4x}+c_{2}e^{2x}+2e^{3x}

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y^{''}-6y^'+8y=-2e^{3x}

y^{^{''}} - 6 y^{^{'}} + 8 y = - 2 e^{3 x}

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

Solution steps

y^{''} (x) - 6 y^{'} (x) + 8 y = - 2 e^{3 x}

Solve linear ODE:

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

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

y^{^{''}} + 5 y^{^{'}} - 5 y = 5 t + 9

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

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

y^{^{'''}} - 9 y^{^{''}} + 27 y^{^{'}} - 27 y = e^{3 x}

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

What is the solution for y^{''}-6y^'+8y=-2e^{3x} ?
The solution for y^{''}-6y^'+8y=-2e^{3x} is y=c_{1}e^{4x}+c_{2}e^{2x}+2e^{3x}