What is the solution for y^{''}+64y=0,y(0)=4,y^'(0)=7 ?

The solution for y^{''}+64y=0,y(0)=4,y^'(0)=7 is y=4cos(8t)+7/8 sin(8t)

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

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

y = 4 cos (8 t) + \frac{7}{8} sin (8 t)

Solution steps

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

Solve linear ODE:

y = 4 cos (8 t) + \frac{7}{8} sin (8 t)

y = 4 cos (8 t) + \frac{7}{8} sin (8 t)

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

y^{^{''}} + 10 y^{^{'}} + \frac{8}{3} y = 0

y^{^{''}} - 2 y^{^{'}} + 2 y = 0, y (\frac{π}{2}) = 0, y (π) = - 1

What is the solution for y^{''}+64y=0,y(0)=4,y^'(0)=7 ?
The solution for y^{''}+64y=0,y(0)=4,y^'(0)=7 is y=4cos(8t)+7/8 sin(8t)