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

The solution for y^{''}-2y^'=6x+6-2e^{4x} is y=c_{1}e^{2x}+c_{2}+(-3x^2-9x)/2-1/4 e^{4x}

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

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

y = c_{1} e^{2 x} + c_{2} + \frac{- 3 x ^{2} - 9 x}{2} - \frac{1}{4} e^{4 x}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} + \frac{- 3 x ^{2} - 9 x}{2} - \frac{1}{4} e^{4 x}

y = c_{1} e^{2 x} + c_{2} + \frac{- 3 x ^{2} - 9 x}{2} - \frac{1}{4} e^{4 x}

y^{^{''}} - 4 y^{^{'}} + 4 y = 15 sin (x) - 35 cos (x)

y^{^{''}} + 9 y = cos (2 t), y (\frac{π}{2}) = - 1, y (0) = 1

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

y^{^{''}} - a^{2} \cdot y + b = 0

y^{^{''}} - 4 y = 8 e^{2 x}, y (0) = 1, y (0) = 0

What is the solution for y^{''}-2y^'=6x+6-2e^{4x} ?
The solution for y^{''}-2y^'=6x+6-2e^{4x} is y=c_{1}e^{2x}+c_{2}+(-3x^2-9x)/2-1/4 e^{4x}