What is the solution for y^{''}+64y=0,y(0)=2,y^'(0)=1 ?

The solution for y^{''}+64y=0,y(0)=2,y^'(0)=1 is y=2cos(8t)+1/8 sin(8t)

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y^{''}+64y=0,y(0)=2,y^'(0)=1

y^{^{''}} + 64 y = 0, y (0) = 2, y^{^{'}} (0) = 1

y = 2 cos (8 t) + \frac{1}{8} sin (8 t)

Solution steps

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

Solve linear ODE:

y = 2 cos (8 t) + \frac{1}{8} sin (8 t)

y = 2 cos (8 t) + \frac{1}{8} sin (8 t)

(- x y sin (x) + 2 y cos (x)) d x + (2 x cos (x)) d y = 0

t^{2} e^{- 3 t} sin (4 t)

\frac{d}{d x} ((\frac{x ^{2} + 4}{x ^{2} - 4})^{3})

\int \frac{2 x + 3}{4 x ^{2} - 9} d x

What is the solution for y^{''}+64y=0,y(0)=2,y^'(0)=1 ?
The solution for y^{''}+64y=0,y(0)=2,y^'(0)=1 is y=2cos(8t)+1/8 sin(8t)