University:
Whitman College
Course:
MATH-244 | Differential Equations
Academic year:

2024
Views:

46

Pages:

5
Author:

arterie4b8

Solutions: Section 2.4 1. Problems 1-6 ask you to apply the Existence and Uniqueness Theorem to a given linear ODE. Be sure to put the DE in standard form first! Some notes as you do this: • The interval is a single (connected) interval. • For theoretical reasons, our interval should be open (so it is possible to differentiate the function at each point in the domain). • The actual interval may be larger than the one guaranteed by the theorem (but we are looking for the one guaranteed by the theorem). For example, in Exercise 5: (4 − t2 )y 0 + 2ty = 3t2 y(1) = −3 y0 + ⇒ 2t 3t2 y= (2 − t)(2 + t) 4 − t2 The functions p, q are continuous on (−∞, −2) (−2, 2) (2, ∞). We want the interval containing t = 1, which is the middle interval. S S 2. Exercises 7-12 ask you to apply the general existence and uniqueness theorem. Sometimes it can be difficult to tell where a functions in the plane are continuous (especially if we had to use the definition), but we are looking for common constraints, like where the denominator is zero, or making sure the log or square root are defined. • Problem 7: f (t, y) = t−y 2t + 5y fy = −7t (2t + 5y)2 ) Therefore, f and fy are both continuous for all (t, y) except those points along the line 2t + 5y = 0, or y = −2t/5 • Problem 9: f (t, y) = ln |ty| 1 − t2 + y 2 fy = (1/y)(1 − t2 + y 2 ) − ln |ty|(2y) (1 − t2 + y 2 )2 Therefore, the derivative did not add any new restrictions- f is continuous on the (t, y) plane except on the axes t = 0 and y = 0, and the curve 1 − t2 + y 2 = 0. 3. In problems 13-16, we solve the differential equation to determine the full interval on which solutions exist (and how they depend on the initial condition). We show in detail the solution to Exercise 13 and 14: • 2.4, 13: Given y 0 = − 4ty , we see that f (t, y) from the Existence and Uniqueness theorem is 4t 4t f (t, y) = − fy (t, y) = 2 y y so the Existence and Uniqueness theorem holds as long as y(0) = y0 6= 0. Now, solving the IVP (it is separable): y dy = −4t dt ⇒ 1 2 y = −2t2 + C1 2 ⇒ y 2 = −4t2 + C2 Therefore, solving for the C2 , we get C2 = y02 , and q y(t) = ± y02 − 4t2 This is valid as long as y > 0 and the expression under the radical sign is positive: y02 − 4t2 > 0 ⇒ 4t2 < y02 ⇒ t2 < y02 4 This is true as long as (assuming y0 > 0): − y0 y0 0, choose the positive root (Graphically, this is the upper or lower half of the ellipse 4t2 + y 2 = y02 ). • 2.4, 14 Given y 0 = 2ty 2 , y(0) = y0 , we see it is separable: Z y −2 dy = Z 2t dt ⇒ 1 − = t2 + C y With the initial condition, C = −1/y0 , so: y(t) = 1 y0 = (1/y0 ) − t2 1 − y0 t2 This is valid as long as y0 6= 0. What if it is? Then we see that y(t) = 0 is the unique solution. If y0 6= 0, then we continue by looking at where 1 − y0 t2 = 0 ⇒ 1 t = ±√ y0 This is valid only if y0 > 0. If y0 < 0, then the denominator, 1 − y0 t2 is never zero (for any t). Thus, if y0 < 0, the solution that we previously obtained is valid for all t. The last case is where y0 > 0. Since the initial time is t0 = 0, then the solution y(t) is only valid for: 1 1 −√ < t < √ y0 y0 Summary: In this homework problem, we saw that the time interval on which the solution is valid depended greatly on the initial value of y, – If y0 < 0, y(t) is valid for all time. – If y0 = 0, y(t) = 0 is the solution, valid for all time. √ – If y0 > 0, y(t) is valid for a short segment of time, between ±1/ y0 . 4. Exercises 21-22 have solutions in the back of the text, but here is 22: 5. 2.4, 22: Part (a) involves substituting the expressions in to be sure the DE is satisfied. For part (b), we note that t2 + 4y ≥ 0 for the square root. In particular, if y = 1 − t, then this implies that 1 1 − t ≥ − t2 4 which is true for all t (graph them), and the second function is clearly valid. Using the existence and uniqueness theorem, we see that Fy (t, y) = √ 1 , which does t2 +4y not exist at (2, −1). The point of part (c) is to show that the second solution, y2 (t) cannot be found as a constant multiple of the first (why? This will become clearer in Chapter 3). Clearly, no value of a constant in ct + c2 can give you the function − 14 t2 . 6. Exercises 23-25: If you have had linear algebra, you might see that there is an underlying point to these- Linearity. If you have not, don’t worry about it yet. However, some students are not sure about what constitutes an answer, so they are provided below: • Problem 23: (a) Show that e2t and ce2t (c any constant) are both solutions to the ODE: y 0 − 2y = 0. You can show this directly (by substitution), or by actually solving the DE. You should see that the general solution is y(t) = Ae2t (b) Show that 1t is a solution to y 0 + y 2 = 0, but Ct is not. You can again show this directly (by substitution), or by actually solving the DE. If you solve it, you should get: y(t) = (or y(t) = 0). 1 t−C • Problem 24: To show this, first note that, if y(t) = φ(t) is a solution to y 0 +p(t)y = 0, then: φ0 + p(t)φ = 0 Now substitute y(t) = cφ: y 0 = cφ0 , and: cφ0 + p(t)cφ = c (φ0 + p(t)φ) = c · 0 = 0 • Problem 25: Same idea as 24. Substitute the expression in to see what you get. Assume that y1 solves y 0 + p(t)y = 0. This means that y10 + p(t)y10 = 0. Assume that y2 solves y 0 + p(t)y = g(t). This means that y20 + p(t)y2 = g(t). Now, substitute y = y1 + y2 , y 0 = y10 + y20 into the DE: (y10 + y20 ) + p(t)(y1 + y2 ) = (y10 + p(t)y1 ) + (y20 + p(t)y2 ) = 0 + g(t) = g(t) 7. Exercises 27-31 focus on a class of DE’s known as Bernoulli equations. Exercise 27 steps you through the process: y 0 + p(t)y = q(t)y n This is “almost” a linear DE- Let v = ⇒ 1 y n−1 y0 1 + p(t) n−1 = q(t) n y y = y 1−n . Then v 0 = (1 − n)y 1−n−1 y 0 = (1 − n) y0 yn Therefore, if we multiply both sides by 1 − n, then we can substitute: y0 1 (1 − n) n + (1 − n)p(t) n−1 = (1 − n)q(t) y y ⇒ v 0 + (1 − n)v = (1 − n)q(t) which is a linear DE in v. Now we’ll use this technique to solve Exercise 28, and 29 is similar: Given t2 y 0 + 2ty = y 3 , divide by t2 y 3 to get the equation in a form we can use: y0 2 1 1 + = 2 3 2 y ty t And we’ll substitute: v = y −2 , so that v 0 = −2y −3 y 0 . To get the substitution, multiply the DE by −2: y0 4 1 2 4 2 0 −2 3 − = − ⇒ v − v = − y t y2 t2 t t2 Now the integrating factor is t−4 , so that 2 + Ct4 v(t) = 5t ⇒ 1 2 + C2 t5 = y2 5t s ⇒ y(t) = ± 5t 2 + C2 t 5 8. Exercise 33 gives us a chance to work with discontinuities (we will pick these up again in Chapter 6, Laplace Transforms). We can solve it now, though. In this case, p(t) depends on time, so we can solve it in pieces: y 0 + 2y = 0 y(0) = 1 0 ≤ t ≤ 1 And y 0 + y = 0, t > 1 If we solve each of these, we get: ( y(t) = e−2t P e−t if 0 ≤ t ≤ 1 if t > 1 where P is any constant. If it is possible to find P so that y is continuous at all time, then we should go ahead and note that: For y to be continuous, we must have: e−2(1) = P e−1 ⇒ P = e−1 Therefore, for t > 1, we can write y as e−1 e−t = e−(t+1)

Differential Equations Solutions Section 2.4 - Study Guide

Related Documents

Description

Related Documents

Free up your schedule!

Take 5 seconds to unlock