In this problem, x = c1 cos(t) + c2 sin(t) is a two-parameter family of solutions of the second-order DE x'' + x = 0. Find a solution of the second-order IVP consisting of this differential equation and the given initial conditions. x(π/2) = 0, x'(π/2

Incomplete Question

The value of x'(π/2) is not given.

I'll assume x'(π/2) to be 1

x(t) = cos(t)

Step-by-step explanation:

Given

x(t) = c1 cos(t) + c2 sin(t)

x(π/2) = 0

x'(π/2) = 1

First, we'll differentiate x(t) to give x'(t)

x'(t) = - c1 sin(t) + c2 cos(t)

Given that x(π/2) = 0

We substitute π/2 for t in x(t)

So,

x(t) = c1 cos(t) + c2 sin(t) becomes

c1 cos(π/2) + c2 sin(π/2) = 0

cos (π/2) = 0 and sin(π/2) = 1;

So, we have

c1 * 0 + c2 * 1 = 0

0 + c2 = 0

c2 = 0.

Given that x'(π/2) = 1

We substitute π/2 for t in x'(t).

So,

x'(t) = - c1 sin(t) + c2 cos(t) becomes

-c1 * sin(π/2) + c2 * cos(π/2) = 1

-c1 * 1 + c2 * 0 = 1

-c1 =+ 0 = 1

-c1 = 1

c1 = 1.

So,

x(t) = c1 cos(t) + c2 sin(t)

=>

x(t) = 1 * cos(t) + 0 * sin(t)

x(t) = cos(t)

7

step-by-step explanation:

yuo ned the activity or vertical angle of the form of 1,7 but 2,2 has a c4,2 as it verticie so i not have a car

step-by-step explanation:

part a. the differences are decreasing, so we know the sequence is not arithmetic. the ratios are constant (=3/4), so we know the sequence is geometric.

part b. from part a, we know the common ratio is 3/4, so each term is 3/4 of the one before. the initial term is the number of gallons at h=1.

g[h] = (3/4)·g[h-1]

g[1] = 384 . . . . initial value required for a recursive definition

part c. the general formula for a geometric sequence is

a[n] = a[1]·r^(n-1)

we have a[1] = 384, r = 3/4, and the term number given by h. the formula that follows is

g[h] = 384·(3/4)^(h-1)

part d. filling in the value and doing the arithmetic, we have

g[12] = 384·(3/4)^11 ≈ 384·0.042235136 ≈ 16.2183 . . gallons

part e. each gallon is 4 quarts, so

16.2183 gal × 4 qt/gal = 64.8732 qt