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

The solution for y^{''}+9y=2x^2-5 is y=c_{1}cos(3x)+c_{2}sin(3x)+(2x^2)/9-49/81

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

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

y = c_{1} cos (3 x) + c_{2} sin (3 x) + \frac{2 x ^{2}}{9} - \frac{49}{81}

Solution steps

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

Solve linear ODE:

y = c_{1} cos (3 x) + c_{2} sin (3 x) + \frac{2 x ^{2}}{9} - \frac{49}{81}

y = c_{1} cos (3 x) + c_{2} sin (3 x) + \frac{2 x ^{2}}{9} - \frac{49}{81}

x \to 1 lim (\frac{x ^{10} - 1}{x - 1})

\frac{d}{d x} (- \frac{2}{( 1 + 2 x ) ^{2}})

\frac{d}{d x} (x e^{2 x})

x \to \infty lim (a^{- x})

\frac{\partial}{\partial x} (\frac{3 x + y}{z})

What is the solution for y^{''}+9y=2x^2-5 ?
The solution for y^{''}+9y=2x^2-5 is y=c_{1}cos(3x)+c_{2}sin(3x)+(2x^2)/9-49/81