What is the solution for y^{''}+9y=2cos(3x) ?

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

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y^{''}+9y=2cos(3x)

y^{^{''}} + 9 y = 2 cos (3 x)

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

Solution steps

y^{''} (x) + 9 y = 2 cos (3 x)

Solve linear ODE:

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

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

\frac{1}{4} e^{3 x}

x \to 1 - lim (x^{2} - 1)

\frac{\partial}{\partial x} (9 x^{5 y})

\frac{d ^{8}}{d x ^{8}} (1 - x^{2})

\frac{3 s}{( s ^{2} + 1 )}

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