What is the solution for y^{''}+16y=0,y(0)=2,y^'(0)=-2 ?

The solution for y^{''}+16y=0,y(0)=2,y^'(0)=-2 is y=2cos(4t)-1/2 sin(4t)

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y^{''}+16y=0,y(0)=2,y^'(0)=-2

y^{^{''}} + 16 y = 0, y (0) = 2, y^{^{'}} (0) = - 2

y = 2 cos (4 t) - \frac{1}{2} sin (4 t)

Solution steps

y^{''} (t) + 16 y = 0

Solve linear ODE:

y = 2 cos (4 t) - \frac{1}{2} sin (4 t)

y = 2 cos (4 t) - \frac{1}{2} sin (4 t)

(9 + 2 x sin (y)) d x - x^{2} cos (y) d y = 0

\frac{d}{d x} (\frac{5}{( x ^{2} + 2 ) ^{3}})

\int e^{\frac{u}{v}} d u

\int (\frac{u - 2}{u}) d u

\int \frac{X ^{2}}{2} d X

What is the solution for y^{''}+16y=0,y(0)=2,y^'(0)=-2 ?
The solution for y^{''}+16y=0,y(0)=2,y^'(0)=-2 is y=2cos(4t)-1/2 sin(4t)