What is the solution for 2y^{''}+3y^'-2y=18x^2-4x-15 ?

The solution for 2y^{''}+3y^'-2y=18x^2-4x-15 is y=c_{1}e^{x/2}+c_{2}e^{-2x}-9x^2-25x-48

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2y^{''}+3y^'-2y=18x^2-4x-15

2 y^{^{''}} + 3 y^{^{'}} - 2 y = 18 x^{2} - 4 x - 15

y = c_{1} e^{\frac{x}{2}} + c_{2} e^{- 2 x} - 9 x^{2} - 25 x - 48

Solution steps

2 y^{''} (x) + 3 y^{'} (x) - 2 y = 18 x^{2} - 4 x - 15

Solve linear ODE:

y = c_{1} e^{\frac{x}{2}} + c_{2} e^{- 2 x} - 9 x^{2} - 25 x - 48

y = c_{1} e^{\frac{x}{2}} + c_{2} e^{- 2 x} - 9 x^{2} - 25 x - 48

x^{^{''}} + x^{^{'}} = sin (t)

y^{^{''}} + 4 y^{^{'}} + 4 y = (5 + x) e - 2 x

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

y^{^{''}} - 4 y^{^{'}} = 8 x e^{3 x}

y^{^{''}} - 6 y^{^{'}} + 9 y = 4 e^{x}

What is the solution for 2y^{''}+3y^'-2y=18x^2-4x-15 ?
The solution for 2y^{''}+3y^'-2y=18x^2-4x-15 is y=c_{1}e^{x/2}+c_{2}e^{-2x}-9x^2-25x-48