What is the solution for y^{''}+64y=0,y(0)=4,y^'(0)=1 ?

The solution for y^{''}+64y=0,y(0)=4,y^'(0)=1 is y=4cos(8t)+1/8 sin(8t)

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y^{''}+64y=0,y(0)=4,y^'(0)=1

y^{^{''}} + 64 y = 0, y (0) = 4, y^{^{'}} (0) = 1

y = 4 cos (8 t) + \frac{1}{8} sin (8 t)

Solution steps

y^{''} (t) + 64 y = 0

Solve linear ODE:

y = 4 cos (8 t) + \frac{1}{8} sin (8 t)

y = 4 cos (8 t) + \frac{1}{8} sin (8 t)

y^{^{''}} + 16 y^{^{'}} + 3 y = 0

y^{^{''}} - 25 y^{^{'}} + 5 y = 0

y^{^{''}} + 10 y^{^{'}} + 25 y = 0, y (1) = 0, y^{^{'}} (1) = - 1

4 y^{^{''}} + 75 y = 0

x^{^{''}} + 5 x^{^{'}} + 9 x = 0

What is the solution for y^{''}+64y=0,y(0)=4,y^'(0)=1 ?
The solution for y^{''}+64y=0,y(0)=4,y^'(0)=1 is y=4cos(8t)+1/8 sin(8t)