Lecture 4: Exact ODE s

Transcription

1 Lecture 4: Exact ODE s Dr. Michael Doughert Januar 23, 203 Exact equations are first-order ODE s of a particular form, and whose methods of solution rel upon basic facts concerning partial derivatives, exact differentials and level curves. These are all Calculus III topics, which we review before introducing the method. The basic idea, which we explain in detail as we proceed, is that we begin with a first-order ODE which we write in the form M(x, )dx + N(x, )d = 0. () We call the ODE () exact if there is a function f(x, ) so that the LHS can be written as df, which b definition is given b df(x, ) = dx + If so, we would have a situation where () could be written The solution would simpl be a one-parameter famil of curves d, or df = f f dx + d. (2) df(x, ) = 0, i.e., (3) f x (x, )dx + f (x, )d = 0. (4) }{{}}{{} M(x,)dx N(x,)d f(x, ) = C, (5) giving as a function of x implicitl. Unfortunatel not all equations can be written (with manageable ease) in the form df = 0, and so we need to be able to test an given ODE with form () to see if it is exact as it stands. We will develop a test for this in Section 7, essentiall solving such ODE s. Before doing so, we will recall partial derivatives in Section, look at the reverse process (integration) in Section 2, define and clarif exact differentials in Section 3, and connect them to level curves in Sections 4 and 5. This will lead to a method for solving exact ODE s which we develop in Sections 7 8. Some further techniques are mentioned in Section 9. The student alread well versed in partial derivatives and differentials can for now look to Section 7 of these notes, though the sections in between ma be valuable for review and context. This makes more sense, perhaps, if we recall from where this came. Suppose we have a function f(x, ), and a path (x(t), (t)), and we want to know how the function changes along the path. According to the chain rule, " # " d f f(x(t), (t)) = dx(t) # f + d(t) dt (x(t),(t)) dt (x(t),(t)) dt = f dx (x(t), (t)) dt + f (x(t), (t)) d dt. The second line is the wa it might appear in a calculus textbook. This might be summarized various was such as df dt = f dx dt + f d dt = f d dt (x, ) = f v, where we assume x and are functions of t, and (x, ) = (x(t), (t)) is some parametrized path. Some texts prefer vector notation, writing the path r (t) = x(t), (t). Its derivative with respect to t is then a velocit vector v. If v =, then f v is the directional derivative of f(x, ) in the direction v. Note that if we multipl the first equation of (2a) in the line above b dt, we get (2), so (2) might be more familiar if we divide it b dt. (2a)

2 Partial Derivatives In this lecture we will assume that we have a function of two variables, namel f(x, ). 2 Such a function will likel var in both x and, so we have partial derivatives to measure these variations: f (x, ) = f x(x, ) = lim x 0 f (x, ) = f (x, ) = lim 0 f(x + x, ) f(x, ), x (6) f(x, + ) f(x, ). (7) Notice that in the first limit (6), the -variable is fixed, i.e., held constant, and the same is true of the x-variable in the second limit (7). If we hold fixed, then f(x, ) becomes a function of x alone, and f x measures how f changes with x. Similarl if we hold x fixed, then f(x, ) becomes a function of alone, with f measuring how it changes with. Computing these partial derivatives is just a matter of treating the other, fixed variable as a constant and using the usual differentiation rules. Example If f(x, ) = sinx e x2, then to compute f/ we hold constant and take the derivative with respect to x. Similarl to find f/ we take x to be constant and differentiate with respect to. Thus [ [f(x, )] = sin x e x2] = sin x = (sin x)e x2 [ e x2] + e x2 [sin x] [ x 2 ] + e x2 cosx = (sin x)e x2 ( 2 ) + e x2 cosx = e ( x2 2 sin x + cosx ), [f(x, )] = [ sin x e x2] = sin x e [ x2 x 2 ] = sin x e x2 x 2 = 2xe x2 sin x. To find f/, we treat no differentl than an other constant, like 5 or π or 0 6. Similarl when the roles of x and are reversed. Since the partial differential operator / treats as a constant, and / treats x as a constant, we have 3 () = 0, (8) (x) = 0. (9) Later it will be important to note that if we have a function of onl, or of x onl, we likewise have 4 ψ(x) = ψ (x), φ() = 0, ψ(x) = 0, (0) φ() = φ (). () 2 Much of what we do here will also work for more variables, i.e., functions f(x,, z) and so on. 3 Note that / is ver different from d/dx. This is because with partial derivatives we assume x and are both independent variables, as in z = f(x, ), while with ordinar derivatives we assume is a function of x (at least locall), i.e., = (x). The distinction is ver important, and often a source of errors even among advanced mathematics students. 2

3 This is because φ() does not change with x, and ψ(x) does not change with. For example, the following ma first appear to be ugl calculations, but reall are quite trivial because of what the partial differential operators consider constant in each case: [ ( 2 + )e sin ] sec 2 + = 0, [ x 2 + 9x ln(x 2 ] + 5) 3 = 0. x4 9x Example 2 Calculate f/ and f/ if: f(x, ) = xsin + x cosx + tan. Solution: Each computation is shortened b recalling which variable is held constant in each case: = [ xsin + x cosx + tan ] = sin + 3x 2 2 sinx, = [ xsin + x cosx + tan ] = xcos + x sec 2. 2 Partial Derivatives and Integration In this section we just point out one ke technical point which arises when we tr to find f(x, ) given one of its partial derivatives: what constitutes a constant of integration when multiple variables are involved. For instance, suppose we know that f x (x, ) = xcosx 2. Since this is a statement about taking a derivative of f and getting xcosx 2, it is reasonable that we somehow take an antiderivative to recover the function, at least as far as we can. For this case, we get = xcosx 2 = f(x, ) = xcosx 2 dx = 2 sin x2 + φ(). (2) This is the most general antiderivative, in x, of xcosx 2. Rather than a simple +C as we need in single-variable calculus, the general term with zero x-partial derivative is an function of alone. A quick check verifies that [ 2 sin x2 + φ() ] = ( cosx 2 ) 2x + 0 = xcos x 2, 2 i.e., the same as our given f x, as desired. If we are given more data about f, we ma be able to determine exactl the form of φ() (just as we were often able to determine the value of the C in single-variable calculus from added data). Note that constants are also functions of (albeit trivial ones), so the solution (2) does include the +C case, built into φ(). Also note how acted like a constant throughout in both calculations. Below are some more sample integrations which illustrate these principles. e x dx = ex + φ(), e x d = x ex + ψ(x), (x ) dx = 2 x2 2 6x + 9x + φ(), dx = x + φ(), 2 2 (x ) d = 3 x ψ(x), d = 2 sin + ψ(x). 4 Aside from the philosophical reasons given in the paragraphs for the equations φ()/ = 0 and ψ(x)/ = 0 from (0) and () namel that the functions do not var with the variable of differentiation we can also see that both follow from the chain rule: φ() = φ () = φ () 0 = 0, ψ(x) = ψ (x) = ψ (x) 0 = 0. 3

4 When dealing with definite double and triple integrals (with endpoints, also known as limits of integration ) in Calculus III, the extra functions φ() or ψ(x) were usuall ignored because the disappeared in exactl the same wa the +C disappears with definite single integrals. However, for this lecture and the method presented here, we will need to pa closer attention to these extra functions, to eventuall find the specific form needed for the solution of the original ODE. 3 Exact Differentials If we are given a function f(x, ) of two variables, we define the exact differential, or total differential of f b df = f f dx + d. (3) As mentioned in Footnote, this is justified when we consider the case in which x and are both functions of a parameter, sa t. If so, then (x, ) = (x(t), (t)), making f(x, ) = f(x(t), (t)) ultimatel a function of t, at least along the path traced b (x, ) = (x(t), (t)). Then we can take the t-derivative, using the following Calc III-tpe chain rule: df(x, ) dt = dx + d dt dt. (4) Hence the chain rule result (4) is the absolute differential (3) divided b dt. Put another wa, we get (3) b multipling (4) b dt. What is interesting about the exact differential (3) is that its form does not depend upon the path; we can calculate its form from that of f without reference to an parametrization. Clearl (4) depends upon the particular parametrization of the path running through a given point (x, ), while (3) does not. 5 Example 3 Below are some simple calculations of exact differentials. (a) If f(x, ) = x 2 + 3, then df = 2xdx d. (b) If f(x, ) = sin xcos, then df = cosxcos dx sin xsin d. 4 Level Curves and Total Differentials Recall that man curves in the x-plane are given as level curves of functions, and are thus written implicitl as f(x, ) = C. If we parametrize such a curve, sa x = x(t), = (t), where x and are differentiable, then for all t in the domain of x(t) and (t) we have f(x(t), (t)) = C. (5) Thus the LHS of (5) is constant in t, and will have t-derivative zero. Appling d dt to both sides of (5) gives us d [f(x(t), (t))] = 0. (6) dt If f(x, ), x(t) and (t) are differentiable, we can expand the LHS of (6) using the chain rule (4) which gives us, for all relevant t, f(x(t), (t)) dx f(x(t), (t)) + d = 0. (7) dt dt Now this is true regardless of the curve s parametrization, b (5). Multipling b dt we get dx + d = 0. (8) 5 As an aside, we can use df in (3) to approximate a small change f in f when we change x b dx and b d, which is a technique found in Calculus I for single-variable functions. 4

5 We recognize LHS of (8) as a total, i.e., exact differential of f (as introduced in (3)) and so (8) becomes df(x, ) = 0. (9) 5 Zero Total Differentials and Level Curves A kind of converse of (9) is the following. If we are given an ODE of the form df = 0, then the solution will be of the form f(x, ) = C, which is a one-parameter famil of implicit curves. To highlight this we will tag it as a separate, numbered equation: 6 df(x, ) = 0 = f(x, ) = C. (20) Note that on a path (x(t), (t)) where f(x, ) = C, the derivative d dtf(x(t), (t)) will be zero, and conversel if such a derivative is zero on such a path (with some differentiabilit conditions), then function should be constant along that path. This is a corollar to the idea from Calculus I that a function is constant if and onl if its derivative is zero, but applied to the variable t here. Thus df = 0 along a level smooth curve (where x(t) and (t) have continuous first derivatives) is the same as f(x, ) = C on that curve. 6 Note on Notation and Higher-Order Partial Derivatives We have used two different notations for partial derivatives. For instance, f x (x, ) =, f (x, ) =. Both notations can be seen as appling a partial differential operator to f. Unfortunatel, the notations are different in the orders that the variables of differentiation appear in the differential operations, which is due to locations of the notations for the operations. For instance, f x (x, ) = (f(x, )) x and f (x, ) = (f(x, )). Contrast this with the other, Leibniz-stle notation, where the variable of differentiation appears on the left of the function: = [f(x, )], and = [f(x, )]. The usual place for confusion is with the higher-order mixed partial derivatives. For instance, f x (x, ) = (f x (x, )) = [ ] = 2 f(x, ). In both cases, we take the partial derivative with respect to x first, and then take the partial derivative of the result with respect to, but the order in which x and appear differs. Recall from Calculus III that if f(x, ) has all continuous second-order partial derivatives f xx, f, f x and f x, then the mixed partial derivatives are the same: 6 As an aside, notice that we can get from (8) an equation for the slope d/dx along that level curve: d = dx = d dx = f/ f/. Man calculus textbooks write this using the counter-intuitive formula d/dx = f x/f, correct in the context of slopes of level curves given b the formula f(x, ) = C. The onl reason to mention this here is that, for some cases, we can take a problem about d/dx, work backwards algebraicall and turn it into a question about total differentials. 5

6 Theorem If R = {(x, ) : x (a, b), (c, d)} is an open rectangle in the x plane, then f xx continuous in R f continuous in R = f x = f x in R. (2) f x continuous in R f x continuous in R Most of the functions we deal with are continuous and have nice, continuous partial derivatives of all orders, except where undefined or otherwise obviousl misbehaving. The following ver simple example shows this theorem in action. A more extensive collection can be found in most calculus textbooks. Example 4 Suppose f(x, ) = e x2 sin. Then f x (x, ) = e x2 2xsin = f x (x, ) = (f x (x, )) = e x2 2xcos; f (x, ) = e x2 cos = f x (x, ) = (f (x, )) x = e x2 2xcos. 7 Testing for Exactness and Solving Exact ODE s Now we collect everthing from before and appl it to some ODE s. First we will look at the abstract problem, and then we will do specific examples. Suppose we are given an ODE of the form M(x, )dx + N(x, )d = 0. (22) To test if the LHS is exact, i.e., can be written df, we identif what corresponds to what. Recall df = f f dx + d. (23) Then (from Linear Algebraic considerations regarding Vector Spaces, but reasonable enough on its face) we must have M(x, ) = f/, and N(x, ) = f/: M(x, ) }{{} f/ dx + N(x, ) d = 0. }{{} f/ If f has continuous second partials, i.e., M and N have continuous first partials, then f x = f x, i.e., [ ] f(x, ) = [ ] f(x, ) (24) becomes or M(x, ) = N(x, ), (25) M = N x. If M = N x, finding f is a matter of careful integration, and quite doable. For now it is important to note that such an f(x, ) will exist if M = N x, and vice versa. A summar of this thinking is contained in the following theorem, which can be found in some form in most three-semester calculus textbooks. Theorem 2 If M(x, ) and N(x, ) are continuous, along with their first partial derivatives, in some rectangular region {(x, ) : x (a, b), (c, d)}, then M(x, )dx + N(x, )d = df(x, ), for some function f M = N. (26) 6

7 Thus M = N x is necessar and sufficient for the existence of some f(x, ) so that df = M dx+n d. The mixed partial derivatives of f, namel f x and f x, are then just M and N x, respectivel, and should be equal. If M = N x, then (22), namel M dx+n d = 0, is reall of the form df(x, ) = 0, with solution f(x, ) = C. If indeed the equation (22) is exact, which we test b seeing if M = N x, then the next challenge is to find f(x, ) so that df = M dx + N d. This is not difficult, but requires some care. It is not difficult because we know, for instance that M(x, ) = f x (x, ), and so f(x, ) = M(x, )dx, (27) except that our constant of integration is then some φ(). To find φ(), we then use the fact that N(x, ) = f (x, ) = f(x, ). (28) Thus we just need to integrate M with respect to x as in (27), and reconcile the resulting antiderivative s form with the notion that its -partial derivative will be N(x, ), i.e., (28). This technique for solving an exact equation is not difficult, though the logic which leads us to a solution is somewhat complicated. We can see the analsis in action in the examples below. Example 5 Consider the ODE 2xsin dx + ( 3 + x 2 cos)d = 0. Here M(x, ) = 2xsin and N(x, ) = 3 + x 2 cos. To test for exactness we compute: (think (f x ) ) M(x, ) = 2xcos = M = N x ODE is exact. (think (f ) x ) N(x, ) = 2xcos Since the ODE is exact, we can go about finding the function f(x, ) so that df = dx + d = 2xsin dx + ( 3 + x 2 cos) d = 0. }{{}}{{} M N Since f/ = M(x, ) = 2xsin, we have f(x, ) = 2xsin dx, f(x, ) = x 2 sin + φ(). Now we use this working form of f(x, ) and the fact that f/ = N(x, ) to find φ(): f i.e., = N(x, ) = x 2 cos + φ () = 3 + x 2 cos = φ () = 3 (29) = φ() = 3 d = 4 4 We usuall do not bother writing the +C (just as we do not write the arbitrar constant in going from dv to v in integration b parts) because its effect is alread present in what will be the final answer as we will see in a moment. 7

8 At this point, we have our form f(x, ) = x 2 sin + φ() becomes f(x, ) = x 2 sin Recall that the ODE was found to be of the form df = 0, so the final answer is f(x, ) = C, i.e., x 2 sin = C. Note that the arbitrar constant of integration for φ ()d = φ() + C can be absorbed into the parameter C on the RHS of the solution to the ODE, so we do not write it when we compute φ. As we work more examples, we will somewhat abbreviate the method to exclude some explanations. One point worth noting is that our method should ield φ () to be a function of onl in (29), and so if x cannot cancel in that line, there must have been some error. ( ) Example 6 Solve if possible: (2x + 2 e x 6 sin x)dx + x 2 + 2e x + 2 d = 0. Solution: Here M = 2x + 2 e x 6 sinx and N = x 2 + 2e x + 2. We next test for exactness: M = [ 2x + 2 e x 6 sinx ] = 2x + 2e x N x = [ x 2 + 2e x + ] the same, ODE is exact 2 = 2x + 2e x Now f(x, ) = M dx = (2x + 2 e x 6 sinx)dx = x e x + 6 cosx + φ(). Using f = N we then get x 2 + 2e x + φ () = x 2 + 2e x + 2, and so φ () = 2 meaning φ() = suffices. Our function f(x, ) is thus f(x, ) = x e x + 6 cosx +, and the solution of our original ODE is thus f(x, ) = C, or x e x + 6 cosx + = C. An alternative to computing f(x, ) = M(x, )dx would be to compute f(x, ) = N(x, )d. Either wa we would have to compute the extra additive term b plaing M and N off of each other. For the example above, we could have computed ( f(x, ) = N d = x 2 + 2e x + ) 2 d = x e x + + ψ(x), and then used f x = M: 2x + 2 e x + ψ (x) = 2x + 2 e x 6 sinx = ψ (x) = 6 sinx = ψ(x) = 6 cosx, again getting us f(x, ) = x e x cosx, and our solution being f(x, ) = C, i.e., x e x cosx = C, as before, onl slightl rearranged. 8

9 8 The Method The general method can be quickl outlined, but of course the devil is in the details. Actuall the details are usuall prett routine, once all the parts of the method are kept straight, so at least one devil is also in the organization. The method can be outlined as given below:. Put the equation into the form M(x, )dx + N(x, )d = 0. (Think f x dx + f d = 0, or simpl df = 0. ) 2. Check for exactness: Is M = N x? (This is necessar for f to exist, and just states f x = f x.) If not, the equation is inexact and ou must tr something else. 3. If es, i.e., the equation is exact, do one of the following: Option (a) Integrate M with respect to x. Since f x = M, we have: f(x, ) = M(x, )dx. Be sure to include the constant of integration φ(). Option (b) Integrate N with respect to. Since f = N, we have: f(x, ) = N(x, )d. Be sure to include the constant of integration ψ(x). (The names φ, ψ do not matter, but the forms do. Some texts use g(x) or g(), for instance.) 4. Find the final form of f b using either (a) f = N(x, ) to find the form of φ(); or (b) f x = M(x, ) to find the form of ψ(x). 5. Once the final form of f is found, so the original ODE can be rewritten df = 0, the solution is therefore the one-parameter famil of curves given b f(x, ) = C. (At this point it would be useful to the reader to re-examine the previous example.) Of course, the given equation ma have been separable from the beginning, in which case separation methods would be preferable. In fact, all separable equations are exact, when written M(x)dx + N()d = 0, which gives triviall that M = 0 = N x. However, exact equations are onl occasionall separable so the more general method outlined above is needed for those cases. Example 7 Solve if possible d dx = sin xsec ln cosxsec tan + x. Solution: Here we need to see if we can write this in a form M dx + N d = 0 and if it ields an exact equation. d dx = sinxsec ln cosxsec tan + x ( = cosxsec tan + x ) d = (sin xsec ln) dx ( = (sinxsec ln) dx + cosxsec tan + x ) d = 0 ( = ( sinxsec + ln)dx + cosxsec tan + x ) d = 0. 9

10 We will use this last line as our ODE M dx + N d = 0. Next we check M = [ sin xsec + ln] = sinxsec tan + N x = [ cosxsec tan + x ] = sinxsec tan + = ODE is exact. Continuing, f(x, ) = M(x, )dx = ( sin xsec + ln) dx = cosxsec + xln + φ(). Using f = N we then have cosxsec tan + x + φ () = cosxsec tan + x = φ () = 0. We can take φ() = 0 for this case, and get f(x, ) = cosxsec + xln, and the solution of our ODE is then cosxsec + xln = C. Again, we could have found f b integrating N with respect to, to get a working (tentative) definition of f, and then used f x = M to find the form of the arbitrar function ψ(x) that would have arisen from that computation. 9 Epilogue: Further Techniques It should also be pointed out that not ever first-degree ODE which can be written M dx+n d = 0 is exact, though antime we can solve for d/dx algebraicall we can then multipl b dx and write M dx+n d = 0 (which ma or ma not be exact). However, some such equations are also separable, or first-order linear so our previous techniques can appl. It should also be pointed out that sometimes there is an integrating factor which can make an equation exact. Conversel, if the equation in the above example had been multiplied or divided b another function, the new equation likel would no longer be exact. (In other words, standards for exactness are somewhat exacting.) An interesting discussion of such things can be found in, among other sources, the Schaum s Outline for Differential Equations, or more precisel, Schaum s Outline of Modern Introductor Differential Equations b Richard Bronson. Such things are to some extent art as much as science. For instance, the following are separable but can ield to methods of exact equations.. xdx + d = 0 can be written d(x) = 0 giving x = C, or = C/x. 2. However, what of dx xd = 0? If we multipl b /x 2 we get d(x/) = 0, ielding x/ = C, or = Cx. x d dx x 2 = 0, which is also Of course this theor has been around for centuries in one form or another, and so some interesting techniques have been found for solving other problems of the form M dx + N d = 0. For instance, Schaum s also points out that if N (M N x ) = g(x), i.e., a function of x alone, then exp[ g(x)dx] is an integrating factor. On the other hand, if M (N x M ) = h(), then exp[ h()d] is an integrating factor. In fact we can derive these as we did the integrating factor for + P(x) = f(x), i.e., µ(x) = exp [ P(x)dx ] which would allow us to get (µ) = µf, and solve for easil. So now we see how we could derive such an integrating factor for some inexact equations. First, we would be given M(x, )dx + N(x, )d = 0. 0

11 Multipling b µ we could write µ(x, )M(x, )dx + µ(x, )N(x, )d = 0, (30) which we hope will be exact, and so we hope [µ(x, )M(x, )] = [µ(x, )N(x, )] x, which expands using the product rule to µm + Mµ = µn x + Nµ x. (3) Case : If µ is a function of x onl, then µ = 0 and so (3) becomes µm = µn x + Nµ x = µ(m N x ) = Nµ x = N (M N x ) = µ x x = ln µ = N (M N x )dx [ ] = µ = exp N (M N x )dx. (32) That we do not need the ± case for our final form of µ is again because multipling both sides of (30) b a nonzero constant will not alter the exactness (or lack thereof). Also, the integral will not have a +φ() because the integrand is a function of x onl and is thus an ordinar Calculus I II tpe of integral. If we inserted +C 2 before the exponential, this onl results in another multiplicative constant due to the exponential. (Recall we are looking for an integration factor, not the most general integration factor.) Case 2: If µ is a function of onl, then µ x = 0 and so (3) becomes µm + Mµ = µn x = Mµ = µ(n x M ) µ = µ = M (N x M ) = ln µ = M (N x M )d [ ] = µ = exp M (N x M )d. (33) In fact, we can reverse the analsis above, and conclude the following, for our original ODE (exact or not) in the form M dx + N d = 0:. If N (M N x ) is a function of x onl, then µ(x) = exp [ N (M N x ) dx ] is an integrating factor to make our ODE exact. 2. If M (N x M ) is a function of onl, then µ() = exp [ M (N x M ) d ] is an integrating factor to make our ODE exact. There are other techniques for special cases, but no general technique to solve such equations analticall (which gives exact solutions, as opposed to numericall, which involves approximations).

12 Homework 4-A. Fill out the following (some of which are contained within these notes): (a) = (b) = (c) = (d) = (e) ψ(x) = (f) ψ(x) = (g) φ() = (h) φ() = 2. Calculate f x and f if f(x, ) = ln x Calculate f x and f if f(x, ) = xsin(x 2 ). 4. For f(x, ) = x 2 ln + secx + tan, show that f x = f x where defined (so ou can ignore the domains, etc.). 5. Show that the following equation is exact and solve it: ( e x sin + 3x 2 ) dx + ( e x cos + x ) d = Show that the following equation is exact and solve it: ( 2xe x2 + 2 sec x 2 3x 2) ( ) sec tan dx + + 2e x d = 0. x 7. Algebraicall rewrite the following ODE, showing that it can be put into a form where it is an exact ODE, and solve it: d dx = 2x3 ln 3x x. 8. Consider the following equation (from Polking, Boggess and Arnold, Differential Equations, Second Edition, Pearson/Prentice Hall, 2006, p. 72): (a) Show that this equation is not exact. (x 2)dx + (x 2 x)d = 0. (b) Find an integrating factor using the techniques at the end of Section 9, multiplication b which makes the equation exact. (c) Show that the new equation is in fact exact. (d) Solve the new, exact ODE. Your answer should look something like x 2 ln x 2 2 = C. (e) Solve the equation above for, using the quadratic formula or completing the square if necessar. 9. If M dx+n d = 0 is alread exact, but we did not notice that immediatel and went looking for integrating factors, what would our formulas (33) and (32) give us for µ?