What is the solution for y^{''}-5y^'+6y=3x^2+1 ?

The solution for y^{''}-5y^'+6y=3x^2+1 is y=c_{1}e^{3x}+c_{2}e^{2x}+(x^2)/2+(5x)/6+25/36

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y^{''}-5y^'+6y=3x^2+1

y^{^{''}} - 5 y^{^{'}} + 6 y = 3 x^{2} + 1

y = c_{1} e^{3 x} + c_{2} e^{2 x} + \frac{x ^{2}}{2} + \frac{5 x}{6} + \frac{25}{36}

Solution steps

y^{''} (x) - 5 y^{'} (x) + 6 y = 3 x^{2} + 1

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} e^{2 x} + \frac{x ^{2}}{2} + \frac{5 x}{6} + \frac{25}{36}

y = c_{1} e^{3 x} + c_{2} e^{2 x} + \frac{x ^{2}}{2} + \frac{5 x}{6} + \frac{25}{36}

5 y^{^{''}} + 6 y^{^{'}} + 25 y = 3 cos (2 t)

y^{^{''}} - 6 y^{^{'}} + 34 y = 25 e^{3 t}, y (0) = 1

(D^{2} - 9 D + 18) y = e^{e^{- 3 x}}

x^{^{''}} + x = 6 sin (2 t), x (0) = 3, x^{^{'}} (0) = 1

2 y^{^{''}} + y^{^{'}} + 3 y = 33 cos (3 t) - 22 sin (3 t)

What is the solution for y^{''}-5y^'+6y=3x^2+1 ?
The solution for y^{''}-5y^'+6y=3x^2+1 is y=c_{1}e^{3x}+c_{2}e^{2x}+(x^2)/2+(5x)/6+25/36