What is the solution for y^{''}+9y=cos(3x) ?

The solution for y^{''}+9y=cos(3x) is y=c_{1}cos(3x)+c_{2}sin(3x)+(xsin(3x))/6

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y^{''}+9y=cos(3x)

y^{^{''}} + 9 y = cos (3 x)

y = c_{1} cos (3 x) + c_{2} sin (3 x) + \frac{x sin ( 3 x )}{6}

Solution steps

y^{''} (x) + 9 y = cos (3 x)

Solve linear ODE:

y = c_{1} cos (3 x) + c_{2} sin (3 x) + \frac{x sin ( 3 x )}{6}

y = c_{1} cos (3 x) + c_{2} sin (3 x) + \frac{x sin ( 3 x )}{6}

x \to - 8 lim (x (8))

\frac{d}{d x} (\frac{1}{x} - ln (x))

\frac{\partial}{\partial z} (tan (3 + 2 x^{2} y^{3} z^{2}))

(2 x + 1)^{\frac{1}{5}}, at x = 2

n = 0 \sum \infty \frac{n}{( n ^{2} + 3 )}

What is the solution for y^{''}+9y=cos(3x) ?
The solution for y^{''}+9y=cos(3x) is y=c_{1}cos(3x)+c_{2}sin(3x)+(xsin(3x))/6