What is the solution for y^{''}+9y=cos(2x),y(0)=-1,y^'(0)=0 ?

The solution for y^{''}+9y=cos(2x),y(0)=-1,y^'(0)=0 is y= 1/5 cos(2x)-6/5 cos(3x)

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y^{''}+9y=cos(2x),y(0)=-1,y^'(0)=0

y^{^{''}} + 9 y = cos (2 x), y (0) = - 1, y^{^{'}} (0) = 0

y = \frac{1}{5} cos (2 x) - \frac{6}{5} cos (3 x)

Solution steps

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

Solve linear ODE:

y = \frac{1}{5} cos (2 x) - \frac{6}{5} cos (3 x)

y = \frac{1}{5} cos (2 x) - \frac{6}{5} cos (3 x)

y^{^{''}} + 2 y^{^{'}} + 2 y = e^{(- x)} \cdot cos (x)

y^{^{''}} + y - x = 0

y^{^{''}} + 3 y^{^{'}} = x + 3

y^{^{''}} + 3 y^{^{'}} = x^{2} + 9

y^{^{''}} + 9 y = 4 sec (3 x)

What is the solution for y^{''}+9y=cos(2x),y(0)=-1,y^'(0)=0 ?
The solution for y^{''}+9y=cos(2x),y(0)=-1,y^'(0)=0 is y= 1/5 cos(2x)-6/5 cos(3x)