What is the solution for y^{''}+6y^'+13y=13x^2 ?

The solution for y^{''}+6y^'+13y=13x^2 is y=e^{-3x}(c_{1}cos(2x)+c_{2}sin(2x))+x^2-(12x)/(13)+46/169

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y^{''}+6y^'+13y=13x^2

y^{^{''}} + 6 y^{^{'}} + 13 y = 13 x^{2}

y = e^{- 3 x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + x^{2} - \frac{12 x}{13} + \frac{46}{169}

Solution steps

y^{''} (x) + 6 y^{'} (x) + 13 y = 13 x^{2}

Solve linear ODE:

y = e^{- 3 x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + x^{2} - \frac{12 x}{13} + \frac{46}{169}

y = e^{- 3 x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + x^{2} - \frac{12 x}{13} + \frac{46}{169}

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

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

x^{^{''}} - 6 x^{^{'}} + 9 x = 2 t e^{3 t}

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

y^{^{''}} + 2 y^{^{'}} + y = \frac{e ^{- x}}{1 - x ^{2}}

What is the solution for y^{''}+6y^'+13y=13x^2 ?
The solution for y^{''}+6y^'+13y=13x^2 is y=e^{-3x}(c_{1}cos(2x)+c_{2}sin(2x))+x^2-(12x)/(13)+46/169