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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + y^{^{'}} - 56 y = 0, y (0) = 5, y^{^{'}} (0) = 5

0.5 y^{^{''}} + 2 y = 0

4 y^{^{''}} + 8 y^{^{'}} + 5 y = 0

y^{^{''}} + c y^{^{'}} = 0

9 y^{^{''}} - 12 y^{^{'}} + 4 y = 0, y (0) = 2, y^{^{'}} (0) = 3

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