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Unformatted text preview: Nonbook problems 1. Verify that for any nonzero constant b, the function f ( x ) = 1 b cosh( bx ) satisfies the differential equation d 2 y dx 2 b v u u t 1 + dy dx ! 2 = 0 . (Recall that the function “cosh” is defined by cosh( x ) = 1 2 ( e x + e x ).) 2. Verify that on the interval 2 < x < 2, the two continuous functions y ( x ) = √ 4 x 2 and y ( x ) = √ 4 x 2 , obtainable from the implicit solution x 2 + y 2 = 4 of the differential equation x + y dy/dx = 0, are (explicit) solutions of this differential equation. 3. Find the general solution of dy dx = x sin x ln y . 4. Find the general solution of dy dx = tan 1 x ye 2 y . ( Notational reminder : “tan 1 ” denotes the inversetangent function, also called arctangent, and also written “arctan”. It does not denote the reciprocal of the tangent function, which is the cotangent function “cot”.) 5. Consider the initialvalue problem dx dt + x 2 = x, x (0) = x (1) (This is the same DE as in exercise 13 of Section 2.2, which you should do prior to starting this exercise.) For each of the following values of x , find both the solution of the IVP and the domain of the solution. The answers to “What is the domain of the solution?” are given below, but they are not obvious . Do not expect to be able to find a quick way for guessing what the answers will be based on x . To get these answers you will have to solve the IVP first —you should get an explicit formula for x ( t )—and then, from your formula, figure out the domain of the solution (remembering that only intervals are allowed as solutions of ODEs, even if the formula you write down has a larger domain). (a) x = 1 2 Answer for domain: (∞ , ∞ ) (i.e.∞ < t < ∞ ) (b) x = 2 Answer for domain: ( ln2 , ∞ ) (i.e. ln2 < t < ∞ ) (c) x = 2 Answer for domain: (∞ , ln( 3 2 ) (i.e.∞ < t < ln( 3 2 ) ) 1 (d) x = 0 Answer for domain: (∞ , ∞ ) (e) x = 1 Answer for domain: (∞ , ∞ ) 6. Let p be a function that is differentiable on the whole real line, and consider the separable differential equation dy dx = p ( y ) . (2) (Here, the function g ( x ) that you’re used to seeing is just the constant function 1.) Note that if we rename the variables x,t in the previous exercise to y,x respectively, the ODE in that exercise can be rewritten in this form, with p ( y ) = y y 2 . (a) Show that the family of all solutions of (2) is translationinvariant in the following sense: if y = φ ( x ) is a solution on an interval a < x < b , and k is any constant, then y = φ ( x k ) is a solution on the interval a + k < x < b + k . (Said another way: horizontally translating the graph of a solution by any amount, you get the graph of another solution.) (b) Using the Fundamental Existence/Uniqueness Theorem for FirstOrder InitialValue Problems (Theorem 1 on p. 12 in the textbook), show that for every point ( x ,y ) ∈ R 2 , the initialvalue problem dy dx = p ( y ) , y ( x ) = y , (3) has a unique solution on some open interval containing...
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This note was uploaded on 12/15/2011 for the course MAP 2302 taught by Professor Tuncer during the Spring '08 term at University of Florida.
 Spring '08
 TUNCER

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