What is the solution for y^{''}+4y=cos(t),y^'(0)=1,y(0)=1 ?

The solution for y^{''}+4y=cos(t),y^'(0)=1,y(0)=1 is y= 2/3 cos(2t)+1/2 sin(2t)+1/3 cos(t)

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y^{''}+4y=cos(t),y^'(0)=1,y(0)=1

y^{^{''}} + 4 y = cos (t), y^{^{'}} (0) = 1, y (0) = 1

y = \frac{2}{3} cos (2 t) + \frac{1}{2} sin (2 t) + \frac{1}{3} cos (t)

Solution steps

y^{''} (t) + 4 y = cos (t)

Solve linear ODE:

y = \frac{2}{3} cos (2 t) + \frac{1}{2} sin (2 t) + \frac{1}{3} cos (t)

y = \frac{2}{3} cos (2 t) + \frac{1}{2} sin (2 t) + \frac{1}{3} cos (t)

What is the solution for y^{''}+4y=cos(t),y^'(0)=1,y(0)=1 ?
The solution for y^{''}+4y=cos(t),y^'(0)=1,y(0)=1 is y= 2/3 cos(2t)+1/2 sin(2t)+1/3 cos(t)