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

The solution for ay^{''}-2ay^'-3ay=x^2 is y=c_{1}e^{3x}+c_{2}e^{-x}-(x^2)/(3a)+(4x)/(9a)-(14)/(27a)

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ay^{''}-2ay^'-3ay=x^2

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

y = c_{1} e^{3 x} + c_{2} e^{- x} - \frac{x ^{2}}{3 a} + \frac{4 x}{9 a} - \frac{14}{27 a}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} e^{- x} - \frac{x ^{2}}{3 a} + \frac{4 x}{9 a} - \frac{14}{27 a}

y = c_{1} e^{3 x} + c_{2} e^{- x} - \frac{x ^{2}}{3 a} + \frac{4 x}{9 a} - \frac{14}{27 a}

y^{^{''}} - 5 y^{^{'}} + 6 y = 10 sin (2 x) + 2 e^{3 x}

y^{^{''}} - 4 y^{^{'}} + 4 y = 1 - 4 e^{t}

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

y^{^{''}} + 6 y + 9 = 0

y^{^{''}} + 3 y^{^{'}} = 6 x + 5

What is the solution for ay^{''}-2ay^'-3ay=x^2 ?
The solution for ay^{''}-2ay^'-3ay=x^2 is y=c_{1}e^{3x}+c_{2}e^{-x}-(x^2)/(3a)+(4x)/(9a)-(14)/(27a)