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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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

2 y^{^{''}} + 9 y^{^{'}} + 7 y = 0

9 y^{^{''}} - 64 y = 0, y (0) = 1, y^{^{'}} (0) = 0

5 y^{^{''}} + 21 y^{^{'}} + 4 y = 0, y (0) = 1, y^{^{'}} (0) = - 19.2 y

4 y^{^{''}} - y^{^{'}} - 3 y = 0, y (0) = 9, y^{^{'}} (0) = 0.25

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