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

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

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

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

y = c_{1} cos (3 x) + c_{2} sin (3 x) + \frac{3}{5} cos (2 x) + \frac{1}{3}

Solution steps

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

Solve linear ODE:

y = c_{1} cos (3 x) + c_{2} sin (3 x) + \frac{3}{5} cos (2 x) + \frac{1}{3}

y = c_{1} cos (3 x) + c_{2} sin (3 x) + \frac{3}{5} cos (2 x) + \frac{1}{3}

\int 9 - x^{2} (- 2 x) d x

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

\frac{\partial}{\partial y} (x^{2} + y^{2} - 9)

f (x) = 6 x^{2} + x - 3

\frac{t}{2}

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