What is the solution for y^{''}+16y=0,y(0)=3,y^'(0)=1 ?

The solution for y^{''}+16y=0,y(0)=3,y^'(0)=1 is y=3cos(4t)+1/4 sin(4t)

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y^{''}+16y=0,y(0)=3,y^'(0)=1

y^{^{''}} + 16 y = 0, y (0) = 3, y^{^{'}} (0) = 1

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

Solution steps

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

Solve linear ODE:

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

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

What is the solution for y^{''}+16y=0,y(0)=3,y^'(0)=1 ?
The solution for y^{''}+16y=0,y(0)=3,y^'(0)=1 is y=3cos(4t)+1/4 sin(4t)