What is the solution for 5y^{''}+7y^'+3y=x^2+2e^{5x} ?

The solution for 5y^{''}+7y^'+3y=x^2+2e^{5x} is

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5y^{''}+7y^'+3y=x^2+2e^{5x}

5 y^{^{''}} + 7 y^{^{'}} + 3 y = x^{2} + 2 e^{5 x}

y = e^{- \frac{7 x}{10}} (c_{1} cos (\frac{11 x}{10}) + c_{2} sin (\frac{11 x}{10})) + \frac{x ^{2}}{3} - \frac{14 x}{9} + \frac{68}{27} + \frac{2}{163} e^{5 x}

Solution steps

5 y^{''} (x) + 7 y^{'} (x) + 3 y = x^{2} + 2 e^{5 x}

Solve linear ODE:

y = e^{- \frac{7 x}{10}} (c_{1} cos (\frac{11 x}{10}) + c_{2} sin (\frac{11 x}{10})) + \frac{x ^{2}}{3} - \frac{14 x}{9} + \frac{68}{27} + \frac{2}{163} e^{5 x}

y = e^{- \frac{7 x}{10}} (c_{1} cos (\frac{11 x}{10}) + c_{2} sin (\frac{11 x}{10})) + \frac{x ^{2}}{3} - \frac{14 x}{9} + \frac{68}{27} + \frac{2}{163} e^{5 x}

(D^{2} - 2 D + 1) y = cosh (x)

x^{^{''}} + a^{2} x = a^{2} b - c

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

y^{^{''}} - 2 y^{^{'}} - 24 y = 5 e^{3 x} + x^{2}

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

What is the solution for 5y^{''}+7y^'+3y=x^2+2e^{5x} ?
The solution for 5y^{''}+7y^'+3y=x^2+2e^{5x} is