What is the solution for y^{''}+9y=4cos(2x) ?

The solution for y^{''}+9y=4cos(2x) is y=c_{1}cos(3x)+c_{2}sin(3x)+4/5 cos(2x)

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y^{''}+9y=4cos(2x)

y^{^{''}} + 9 y = 4 cos (2 x)

y = c_{1} cos (3 x) + c_{2} sin (3 x) + \frac{4}{5} cos (2 x)

Solution steps

y^{''} (x) + 9 y = 4 cos (2 x)

Solve linear ODE:

y = c_{1} cos (3 x) + c_{2} sin (3 x) + \frac{4}{5} cos (2 x)

y = c_{1} cos (3 x) + c_{2} sin (3 x) + \frac{4}{5} cos (2 x)

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

y^{^{''}} - 4 y = 9 e^{t}, y (0) = 0, y^{^{'}} (0) = 0

x \to - 1 lim (5 - 2 x - x^{2})

\int (2^{x} - 3^{x})^{0} d x

\frac{d}{d x} (7 sec (x) - 2 x)

What is the solution for y^{''}+9y=4cos(2x) ?
The solution for y^{''}+9y=4cos(2x) is y=c_{1}cos(3x)+c_{2}sin(3x)+4/5 cos(2x)