In this problem, y = c1 cos 4x + c2 sin 4x is a two-parameter family of solutions of the second-order de y'' + 16y = 0.
if possible, find a solution of the differential equation that satisfies the given side conditions. the conditions specified at two different points are called boundary conditions. (if not possible, enter not.) y(0) = 0, y(π/8) = 4.

y=4 sin(4x)

Step-by-step explanation:

So you are given y(0)=0.  This means when x=0, y=0.

So plug this in:

0=c1 cos(4*0)+c2 sin(4*0)

0=c1 cos(0)   +c2 sin(0)

0=c1 (1)          +c2 (0)

0=c1               +0

0=c1

So our solution looks like this after applying the first boundary condition:

y=c2 sin(4x).

Now we also have y(pi/8)=4.  This means when x=pi/8, y=4.

So plug this in:

4=c2 sin(4*pi/8)

4=c2 sin(pi/2)

4=c2 (1)

4=c2

So the solution with the given conditions applies is y=4 sin(4x) .

Testing:

y'=16 cos(4x)

y''=-64 sin(4x).

y''+16y=0

-64 sin(4x)+16(4 sin(4x))

-64 sin(4x)+64 sin(4x)

0

So the solution still works.

do i do it on a graph?

step-by-step explanation: