What is the solution for y^{''}+16y=0,y(0)=2,y^'(0)=1 ?

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

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y^{''}+16y=0,y(0)=2,y^'(0)=1

y^{^{''}} + 16 y = 0, y (0) = 2, y^{^{'}} (0) = 1

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

Solution steps

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

Solve linear ODE:

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

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

\frac{\partial}{\partial x} (arcsin (5 x))

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

x \to 1 lim (x ln (x) - x)

\frac{\partial}{\partial x} ((x + y) e^{y})

\int \frac{k}{r} (- \frac{r ^{2}}{4}) d r

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