What is the solution for y^{''}-5y^'+6y=x^2-4 ?

The solution for y^{''}-5y^'+6y=x^2-4 is y=c_{1}e^{3x}+c_{2}e^{2x}+(x^2)/6+(5x)/(18)-53/108

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y^{''}-5y^'+6y=x^2-4

y^{^{''}} - 5 y^{^{'}} + 6 y = x^{2} - 4

y = c_{1} e^{3 x} + c_{2} e^{2 x} + \frac{x ^{2}}{6} + \frac{5 x}{18} - \frac{53}{108}

Solution steps

y^{''} (x) - 5 y^{'} (x) + 6 y = x^{2} - 4

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} e^{2 x} + \frac{x ^{2}}{6} + \frac{5 x}{18} - \frac{53}{108}

y = c_{1} e^{3 x} + c_{2} e^{2 x} + \frac{x ^{2}}{6} + \frac{5 x}{18} - \frac{53}{108}

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

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

y^{^{''}} - 14 y^{^{'}} + 49 y = 10 e^{7 x}

y^{^{''}} + y^{^{'}} - 12 y = sin (t), y (0) = 1, y^{^{'}} (0) = 1

y, (d^{2} + 5 d + 6) y = (d + 1) 6 t^{2}

What is the solution for y^{''}-5y^'+6y=x^2-4 ?
The solution for y^{''}-5y^'+6y=x^2-4 is y=c_{1}e^{3x}+c_{2}e^{2x}+(x^2)/6+(5x)/(18)-53/108