What is the solution for y^{''}+64y=0,y(0)=4,y^'(0)=3 ?

The solution for y^{''}+64y=0,y(0)=4,y^'(0)=3 is y=4cos(8t)+3/8 sin(8t)

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y^{''}+64y=0,y(0)=4,y^'(0)=3

y^{^{''}} + 64 y = 0, y (0) = 4, y^{^{'}} (0) = 3

y = 4 cos (8 t) + \frac{3}{8} sin (8 t)

Solution steps

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

Solve linear ODE:

y = 4 cos (8 t) + \frac{3}{8} sin (8 t)

y = 4 cos (8 t) + \frac{3}{8} sin (8 t)

12 y^{^{''}} - 5 y^{^{'}} - 3 y = 0, y (0) = 2

y^{^{''}} + 64 y = 0, y (0) = 4, y^{^{'}} (0) = 7

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

y^{^{''''}} + 2 y^{^{'''}} + 4 y^{^{''}} + 8 y^{^{'}} = 0

2 \frac{d ^{2} y}{d x ^{2}} + 8 \frac{d y}{d x} - y = 0

What is the solution for y^{''}+64y=0,y(0)=4,y^'(0)=3 ?
The solution for y^{''}+64y=0,y(0)=4,y^'(0)=3 is y=4cos(8t)+3/8 sin(8t)