What is the solution for y^{''}+9y=cos(2t),y(pi/2)=-1,y(0)=1 ?

The solution for y^{''}+9y=cos(2t),y(pi/2)=-1,y(0)=1 is y= 4/5 cos(3t)+4/5 sin(3t)+1/5 cos(2t)

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y^{''}+9y=cos(2t),y(pi/2)=-1,y(0)=1

y^{^{''}} + 9 y = cos (2 t), y (\frac{π}{2}) = - 1, y (0) = 1

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

Solution steps

y^{''} (t) + 9 y = cos (2 t)

Solve linear ODE:

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

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

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

y^{^{''}} - a^{2} \cdot y + b = 0

y^{^{''}} - 4 y = 8 e^{2 x}, y (0) = 1, y (0) = 0

y^{^{''}} + 4 y^{^{'}} = 2 e^{- 4 x} + 3 sin (2 x)

y^{^{''}} - 3 y^{^{'}} + 2 y = cos (t), y (0) = 0, y^{^{'}} (0) = 1

What is the solution for y^{''}+9y=cos(2t),y(pi/2)=-1,y(0)=1 ?
The solution for y^{''}+9y=cos(2t),y(pi/2)=-1,y(0)=1 is y= 4/5 cos(3t)+4/5 sin(3t)+1/5 cos(2t)