What is the solution for S^{''}+16S=0,S(0)=4,S^'(0)=2 ?

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

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S^{''}+16S=0,S(0)=4,S^'(0)=2

S^{^{''}} + 16 S = 0, S (0) = 4, S^{^{'}} (0) = 2

S (t) = 4 cos (4 t) + \frac{1}{2} sin (4 t)

Solution steps

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

Solve linear ODE:

S (t) = 4 cos (4 t) + \frac{1}{2} sin (4 t)

S (t) = 4 cos (4 t) + \frac{1}{2} sin (4 t)

y^{^{''}} - y^{^{'}} - 20 y = 0, y (0) = α, y^{^{'}} (0) = 2

144 y^{^{''}} - 312 y^{^{'}} + 169 y = 0

y^{^{''}} + 25 y = 0, y (0) = 1, y^{^{'}} (0) = - 1

y^{^{''}} - 18 y^{^{'}} + 97 y = 0

\frac{d ^{3} y}{d x ^{3}} + 3 \frac{d ^{2} y}{d x ^{2}} = 0

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