What is the solution for y^{''}+9y=e^{4x}+3x^2-5 ?

The solution for y^{''}+9y=e^{4x}+3x^2-5 is y=c_{1}cos(3x)+c_{2}sin(3x)+1/25 e^{4x}+(x^2)/3-17/27

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y^{''}+9y=e^{4x}+3x^2-5

y^{^{''}} + 9 y = e^{4 x} + 3 x^{2} - 5

y = c_{1} cos (3 x) + c_{2} sin (3 x) + \frac{1}{25} e^{4 x} + \frac{x ^{2}}{3} - \frac{17}{27}

Solution steps

y^{''} (x) + 9 y = e^{4 x} + 3 x^{2} - 5

Solve linear ODE:

y = c_{1} cos (3 x) + c_{2} sin (3 x) + \frac{1}{25} e^{4 x} + \frac{x ^{2}}{3} - \frac{17}{27}

y = c_{1} cos (3 x) + c_{2} sin (3 x) + \frac{1}{25} e^{4 x} + \frac{x ^{2}}{3} - \frac{17}{27}

What is the solution for y^{''}+9y=e^{4x}+3x^2-5 ?
The solution for y^{''}+9y=e^{4x}+3x^2-5 is y=c_{1}cos(3x)+c_{2}sin(3x)+1/25 e^{4x}+(x^2)/3-17/27