What is the solution for y^{''}+g^2y=cos(2t),y(0)=1,y^'(0)=0 ?

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

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y^{''}+g^2y=cos(2t),y(0)=1,y^'(0)=0

y^{^{''}} + g^{2} y = cos (2 t), y (0) = 1, y^{^{'}} (0) = 0

y = \frac{cos ( 2 t )}{g ^{2} - 4} + cos (g t) (- \frac{1}{g ^{2} - 4} + 1)

Solution steps

y^{''} (t) + g^{2} y = cos (2 t)

Solve linear ODE:

y = \frac{cos ( 2 t )}{g ^{2} - 4} + cos (g t) (- \frac{1}{g ^{2} - 4} + 1)

y = \frac{cos ( 2 t )}{g ^{2} - 4} + cos (g t) (- \frac{1}{g ^{2} - 4} + 1)

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

y^{^{''}} - 4 y^{^{'}} - 5 y = 3 e^{- 4 t}

y^{^{''}} + 4 y = 6 cos (2 t), y (0) = 2, y^{^{'}} (0) = 1

y^{^{''}} - 9 y = - 27 t + 18 t^{2}

y^{^{'''}} - 2 y^{^{''}} + 26 y^{^{'}} = 174 e^{3 x}

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