Question Problem 4. (10 POINTS) Consider the following IVP: \[ \left\{\begin{array}{l} y^{\prime \prime}-2 t y^{\prime}+8 y=0 \\ y(0)=4, y^{\prime}(0)=0 \end{array}\right. \] (a) ( 1 point) Is \( t_{0}=0 \) an ordinary point, a regular singular point, or an irregular singular point of the ODE? Justify your answer. (b) (8 points) Use the power series approach to find the solution of the IVP. (c) (1 point) What is the radius of convergence of the resulting power series?

Consider the following differential equation: \( y^{\prime \prime}(x)-y(x)=0 \), where \( x_{0}=0 \) The given differential equation has no singular points; thus, every expansion point \( x_{0} \) is Find the recurrence relation to express \( a_{n+2} \) in terms of \( a_{n} \) and \( n \). Note: Use the underscore symbol to type in "a_n". \[ a_{n+2}=\quad, n \geq 0 \] The recurrence relation tells us that, starting from \( a_{0} \), we can obtain the even-numbered coefficients \( a_{2} \), \( a_{4}, a_{6}, \ldots, a_{2 n} \), where \( n \geq 0 \). Similarly, starting from \( a_{1} \), we can obtain the odd-numbered coefficients \( a_{3}, a_{5}, a_{7}, \ldots, a_{2 n+1} \), where \( n \geq 0 \). - The indices change in steps by , so there exist different solution functions. - The first solution function is of the form \( y_{1}(x)=\sum_{n=0}^{\infty} a_{2 n} x^{p} \), where the power on \( x \) is \( p= \) - The second solution function is given by \( y_{2}(x)=\sum_{n=0}^{\infty} a_{2 n+1} x^{q} \), where the power on \( x \) is \( q= \) Determine the general rule for these coefficients. Assume that \( a_{0} \neq 0 \) and \( a_{1} \neq 0 \). Use factorials to obtain a concise notation. Note: Only the simple \( n \) ! case needs no parentheses; otherwise, we write parentheses before the symbol for factorial. \[ a_{2 n}=a_{0} \] \[ \begin{array}{l} y(x)=a_{0} \sum_{n=2}^{\infty} \frac{x^{2 n}}{(2 n) !}+a_{1} \sum_{n=3}^{\infty} \frac{x^{2 n+1}}{(2 n+1) !} \\ y(x)=a_{0} \sum_{n=0}^{\infty} \frac{x^{n}}{n !}+a_{1} \sum_{n=0}^{\infty} \frac{x^{-n}}{n !} \\ y(x)=a_{0} \sum_{n=0}^{\infty} \frac{x^{n}}{(n+1) !}+a_{0} \sum_{n=0}^{\infty} \frac{x^{2 n}}{(n+2) !} \\ y(x)=a_{0} \sum_{n=0}^{\infty} \frac{x^{2 n+1}}{(n+1) !}+a_{0} \sum_{n=0}^{\infty} \frac{x^{2 n}}{(n+2) !} \\ y(x)=a_{0} \sum_{n=0}^{\infty} \frac{x^{2 n}}{(2 n) !}+a_{1} \sum_{n=0}^{\infty} \frac{x^{2 n+1}}{(2 n+1) !} \end{array} \] None of the above If \( y(x)=a_{0} y_{1}(x)+a_{1} y_{2}(x) \), we can find the two solution functions. Write each solution function in its expanded form, so that the first four terms for each function are shown. \[ \begin{array}{ll} y_{1}(x)= & +\ldots \\ y_{2}(x)= & +\ldots \end{array} \] Finally, investigate the linear independence of the two solution functions. First, evaluate the functions and their derivatives to find \( y_{1}(0), y_{2}(0), y_{1}{ }^{\prime}(0) \), and \( y_{2}{ }^{\prime}(0) \). Use these results to calculate the Wronskian determinant of the two solutions at \( x_{0}=0 \). \[ W\left(y_{1}, y_{2}\right)(0)= \] Select all the true statements. The Wronskian is nonzero on the interval of validity \( (-\infty, \infty) \). The two functions are linearly dependent. The Wronskian is nonzero at \( x_{0}=0 \). The set \( \left\{y_{1}, y_{2}\right\} \) is a fundamental set. The Wronskian is equal to zero at \( x_{0}=0 \). The two solution functions are linearly independent. None of the above.

Lael wants to determine several totals and averages for active students.In cell Q8, enter a formula using the COUNTIF function and structured references to count the number of students who have been elected to offices in student organizations.

A solid sphere of radius 40 cm has a total positive charge of 26 μC uniformly distributed throughout its volume. Calculate the magnitude of the electric field (a) 0 cm, (b) 10 cm, (C) 40 cm and (d) 60 cm from the center of the sphere.

The weights of a certain brand of candies are normally distributed with a mean weight of 0.8602 g and a standard deviation of 0.0521 g. A sample of these candies came from a package containing 456 candies, and the package label stated that the net weight is 389.2 g.​ (If every package has 456 ​candies, the mean weight of the candies must exceed 389.2 / 456 =0.8534 g for the net contents to weigh at least 389.2 ​g.) a. If 1 candy is randomly​ selected, find the probability that it weighs more than 0.8534 g.

Which of the following methods is best to have when a network goes down? O In-band management Site-to-site VPN Client-to-site VPN O Out-of-band management Kate, a network administrator, has been tasked with staying within the company budget. She has a large network and doesn't want to spend more than she needs to on purchasing and registering multiple public IP addresses for each of the hosts on her network. Which of the following methods could help her provide internet access but also keep costs low and limit the number of registered IP addresses her organization needs to purchase? Use Network Address Translation. Use Layer 2 switches 0 Use Layer 3 switches. Use PoE devices

Question 7 of 9 (1 point) | Attempt 1 of Unlimited View question in a popup 6.2 Section Exercise 23 (calc TV sets: According to the Nielsen Company, the mean number of TV sets in a U.S. household in 2013 was 2.24. Assume the standard deviation is 1.4. A sample of 80 households is drawn. Part 1 of 5 B (a) What is the probability that the sample mean number of TV sets is greater than 2? Round your answer to at least four decimal places. இ The probability that the sample mean number of TV sets is greater than 2 is 0.9370 ol. Part 2 of 5 (b) What is the probability that the sample mean number of TV sets is between 2.5 and 3? Round your answer to at least four decimal places. The probability that the sample mean number of TV sets is between 2.5 and 3 is 0.0483 Part 3 of 5 (c) Find the 40th percentile of the sample mean. Round the answer to at least two decimal places. The 40th percentile of the sample mean is 2.20 Part: 3/5 Part 4 of 5 (d) Would it be unusual for the sample mean to be less than 2? Round your answer to at least four decimal places. It is unusual because the probability of the sample mean being less than 2 is Х 5