What is the solution for y^{''}+9y=cos(6t) ?

The solution for y^{''}+9y=cos(6t) is y=c_{1}cos(3t)+c_{2}sin(3t)-1/27 cos(6t)

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y^{''}+9y=cos(6t)

y^{^{''}} + 9 y = cos (6 t)

y = c_{1} cos (3 t) + c_{2} sin (3 t) - \frac{1}{27} cos (6 t)

Solution steps

y^{''} (t) + 9 y = cos (6 t)

Solve linear ODE:

y = c_{1} cos (3 t) + c_{2} sin (3 t) - \frac{1}{27} cos (6 t)

y = c_{1} cos (3 t) + c_{2} sin (3 t) - \frac{1}{27} cos (6 t)

x \to 2 - lim (2)

y = arctan (- 2 x), at x = - 3

\int 14 e^{x} d x

f (x) = 1 - 6 x^{2}, (2, - 23)

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

What is the solution for y^{''}+9y=cos(6t) ?
The solution for y^{''}+9y=cos(6t) is y=c_{1}cos(3t)+c_{2}sin(3t)-1/27 cos(6t)