What is the solution for y^{''}+16y=0,y(0)=0,y^'(0)=2 ?

The solution for y^{''}+16y=0,y(0)=0,y^'(0)=2 is y= 1/2 sin(4t)

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y^{''}+16y=0,y(0)=0,y^'(0)=2

y^{^{''}} + 16 y = 0, y (0) = 0, y^{^{'}} (0) = 2

y = \frac{1}{2} sin (4 t)

Solution steps

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

Solve linear ODE:

y = \frac{1}{2} sin (4 t)

y = \frac{1}{2} sin (4 t)

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

y^{^{''}} - 8 y^{^{'}} + 32 y = 0, y (0) = 5, y^{^{'}} (0) = - 4

3 y^{^{''}} - 11 y^{^{'}} + 2 y = 0

y^{^{''}} - y^{^{'}} = 0, y (0) = 2, y^{^{'}} (0) = 1

x^{^{''}} - 3 x^{^{'}} + 2 x = 0, x (0) = 1, x^{^{'}} (0) = 0

What is the solution for y^{''}+16y=0,y(0)=0,y^'(0)=2 ?
The solution for y^{''}+16y=0,y(0)=0,y^'(0)=2 is y= 1/2 sin(4t)