What is the solution for y^{''}+9y=cos(t),y(0)=0,y^'(0)=0 ?

The solution for y^{''}+9y=cos(t),y(0)=0,y^'(0)=0 is y= 1/8 cos(t)-1/8 cos(3t)

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y^{''}+9y=cos(t),y(0)=0,y^'(0)=0

y^{^{''}} + 9 y = cos (t), y (0) = 0, y^{^{'}} (0) = 0

y = \frac{1}{8} cos (t) - \frac{1}{8} cos (3 t)

Solution steps

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

Solve linear ODE:

y = \frac{1}{8} cos (t) - \frac{1}{8} cos (3 t)

y = \frac{1}{8} cos (t) - \frac{1}{8} cos (3 t)

What is the solution for y^{''}+9y=cos(t),y(0)=0,y^'(0)=0 ?
The solution for y^{''}+9y=cos(t),y(0)=0,y^'(0)=0 is y= 1/8 cos(t)-1/8 cos(3t)