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Note, however, that the coefficients in the differential equations in a and b can be made arbitrarily close by choosing sufficiently small. There are three cases. For Problems 3—6, some details of the calculations are omitted. In Problems 7—16 the method of undetermined coefficients is used to find a particular solution of the nonhomogeneous equation.
Details are included for Problems 7 and 8, and solutions are outlined for the remainder of these problems. This attempt includes both a sine and cosine term even though f x has only a cosine term, because both terms may be needed to find a particular solution. In this case yp x contains only a sine term, although f x has only the cosine term. Finding a particular solution yp x for this problem requires some care.
This will work because neither e2x nor e3x is a solution of the associated homogeneous equation. In Problems 17—24 the strategy is to first find a general solution of the differential equation, then solve for the constants to find a solution satisfying the initial conditions. Problems 17—22 are well suited to the use of undetermined coefficients, while Problems 23 and 24 can be solved fairly directly using variation of parameters.
The solution of the initial value problem is! Details are included with solutions for Problems 1—2, while just the solutions are given for Problems 3— This is a constant coefficient second-order homogeneous differential equation for Y t , which we know how to solve.
We obtain the solution in all cases by solving this linear constant coefficient second-order equation. We know how to solve this problem. The new twist here is that the entire initial value problem including initial conditions was transformed in terms of t and solved for Y t , then this solution Y t in terms of t was transformed back to the solution y x in terms of x.
This is the recurrence relation. These solutions are consistent with the observation that, upon division by 4, the differential equation is an Euler equation. The median is the sixteenth number from the left, or 1 the last 1 to the right in the ordered list. Section 2 Random Variables and Probability Distributions 1. If we roll two dice, there are thirty-six possible outcomes. The sums of the numbers that can come up on the two dice are listed in Table 1.
The table gives all of the values that X o can take on, over all outcomes o of the experiment. Each value is listed as often as it occurs as a value of X. This is interpreted to mean that, on average, we expect to come up with a seven if we roll two dice.
This is a reasonable expectation in view of the fact that there are more ways to roll 7 than any other sum with two dice. The standard deviation is. Flip four coins, with sixteen possible outcomes. If o is an outcome, X o can have only two values, namely 1 if two, three, or four tails are in o, or 3 otherwise one tail or no tails in o.
The values assumed by X are 0, 1, 2, 3, 4. The outcomes of two rolls of the dice are displayed in Table 2. There are 52 C2 ways to do this, disregarding order. If o is an outcome in which both cards are numbered, then X o equals the sum of the numbers on the cards. Now the outcomes have the appearance x, y, T , where x and y can independently be 2, 4 or 6. Section 2 Four Counting Principles 1. The fact that these are letters of the alphabet that we are arranging in order is irrelevant.
The issue is that there are nine distinct ob jects. The number of arrangements is 9! There are 26 letters in the English slphabet. The problem is one of determining the number of ways of choosing 17 ob jects from 26 ob jects, with order taken into account.
This is 26 P The total number of codes is 99 , or 3. These 7 symbols have 7! The number of ways of doing this is 5! For n! There are 12! There are 10 letters from a through l, inclusive.
There are 7! We want to pick 7 ob jects from 25, taking order into account. The number of ways to do this is 25 P7. The number of ballots is 16! Because order is important, the number of possibilities is 22 P6. There are 19 P 2. This is reasonable from a common sense point of view, since, with 20 number to choose from, we would expect 5 percent to begin with any particular one of the numbers. The answer is that 5 percent of the choices ends in 9.
There are 18 such choices. Without order and, we assume, without replacement , the number of ten card hands is 52!
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