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}

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