What is the solution for 16y^{''}+64y=0,y(0)=0.3,y^'(0)=-0.2 ?

The solution for 16y^{''}+64y=0,y(0)=0.3,y^'(0)=-0.2 is y=0.3cos(2t)-0.1sin(2t)

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16y^{''}+64y=0,y(0)=0.3,y^'(0)=-0.2

16 y^{^{''}} + 64 y = 0, y (0) = 0.3, y^{^{'}} (0) = - 0.2

y = 0.3 cos (2 t) - 0.1 sin (2 t)

Solution steps

16 y^{''} (t) + 64 y = 0

Solve linear ODE:

y = 0.3 cos (2 t) - 0.1 sin (2 t)

y = 0.3 cos (2 t) - 0.1 sin (2 t)

What is the solution for 16y^{''}+64y=0,y(0)=0.3,y^'(0)=-0.2 ?
The solution for 16y^{''}+64y=0,y(0)=0.3,y^'(0)=-0.2 is y=0.3cos(2t)-0.1sin(2t)