What is the solution for 64y^{''}+729y=0,y(0)=8,y^'(0)=3 ?

The solution for 64y^{''}+729y=0,y(0)=8,y^'(0)=3 is y=8cos((27t)/8)+8/9 sin((27t)/8)

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64y^{''}+729y=0,y(0)=8,y^'(0)=3

64 y^{^{''}} + 729 y = 0, y (0) = 8, y^{^{'}} (0) = 3

y = 8 cos (\frac{27 t}{8}) + \frac{8}{9} sin (\frac{27 t}{8})

Solution steps

64 y^{''} (t) + 729 y = 0

Solve linear ODE:

y = 8 cos (\frac{27 t}{8}) + \frac{8}{9} sin (\frac{27 t}{8})

y = 8 cos (\frac{27 t}{8}) + \frac{8}{9} sin (\frac{27 t}{8})

y^{^{'''}} + 4 y^{^{''}} - 32 y^{^{'}} = 0

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

0.25 \frac{d ^{2} x}{d t ^{2}} + 0.1 \frac{d x}{d t} + x = 0

\frac{d ^{2} x}{d t ^{2}} + 9.8 x = 0

y^{^{''}} + 0.75 y = 0

What is the solution for 64y^{''}+729y=0,y(0)=8,y^'(0)=3 ?
The solution for 64y^{''}+729y=0,y(0)=8,y^'(0)=3 is y=8cos((27t)/8)+8/9 sin((27t)/8)