y = c1 cos(5x) + c2 sin(5x) is a two-parameter family of solutions of the second-order DE y'' + 25y = 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 IMPOSSIBLE.) y(0) = 1, y'(π) = 9

y = 2cos5x-9/5sin5x

Step-by-step explanation:

Given the solution to the differential equation y'' + 25y = 0 to be

y = c1 cos(5x) + c2 sin(5x). In order to find the solution to the differential equation given the boundary conditions y(0) = 1, y'(π) = 9, we need to first get the constant c1 and c2 and substitute the values back into the original solution.

According to the boundary condition y(0) = 2, it means when x = 0, y = 2

On substituting;

2 = c1cos(5(0)) + c2sin(5(0))

2 = c1cos0+c2sin0

2 = c1 + 0

c1 = 2

Substituting the other boundary condition y'(π) = 9, to do that we need to first get the first differential of y(x) i.e y'(x). Given

y(x) = c1cos5x + c2sin5x

y'(x) = -5c1sin5x + 5c2cos5x

If y'(π) = 9, this means when x = π, y'(x) = 9

On substituting;

9 = -5c1sin5π + 5c2cos5π

9 = -5c1(0) + 5c2(-1)

9 = 0-5c2

-5c2 = 9

c2 = -9/5

Substituting c1 = 2 and c2 = -9/5 into the solution to the general differential equation

y = c1 cos(5x) + c2 sin(5x) will give

y = 2cos5x-9/5sin5x

The final expression gives the required solution to the differential equation.

answer: expected value of x = 1.0

step-by-step explanation:

p(x=0) = 0.25

p(x=1) = 0.50

p(x=2) = 0.25

expected value for x = 0×0.25 + 1×0.5 + 2×0.25 = 1.0