Lecture : Separable Ordinar Differential Equations Dr. Michael Doughert Januar 8, 00 Some Terminolog: ODE s, PDE s, IVP s The differential equations we have looked at so far are called ordinar differential equations, or ODE s, because the involve ordinar derivatives (of an order like, d dp, dt and so on. Another class of differential equations which we are not likel to have time for in this semester is the partial differential equations, or PDE s. Those involve partial derivatives. For instance, Laplace s equation (also known as the equilibrium heat equation is given b u xx + u = 0, where u = u(x,. Heat, wave and electromagnetic theor equations are the most well-known PDE s, but PDE s are ubiquitous in science, engineering and economics. Whenever we have a function (such as temperature, price, etc. of more than one variable the derivatives are necessaril partial derivatives. The techniques for solving such equations are man, varied and different from those for ODE s. The are reserved for the course Differential Equations II, which runs when there is a demand. When we solve an ODE, we are likel to be left with parameters, and therefore a famil of solution curves. This famil of curves, written with parameters, is often called the general solution of the ODE. To determine a specific curve we might need, we usuall specif some data which pins down a (hopefull unique curve satisfing the equation and the data. An equation, together with data to determine a specific curve, is called an initial value problem, or an IVP. Of course we need to know how to solve the underling ODE before we can solve an IVP s. Nonetheless, we can look at a couple of IVP s to see what we are in for there. Example Let us solve the following IVP: = x, (/ = 7. ( The first part is the ODE, and the second is the initial data (technicall datum for this case. Now clearl this is separable, so we continue as before: = = x x = x This is where the terminolog gets fun. Or we could simpl write Z = = x = sin x + C. x = sin x + C.
Now that the ODE is solved, we need onl find the exact curve b finding the C which corresponds to the initial data. For that we just plug in (x, = (/, 7 : 7 = sin (/ + C = π 6 + C 7 π 6 = C. Thus the solution to the IVP is ( = sin x + 7 π. ( 6 Example Consider the IVP = x (8 = 6. (the ODE (initial data (3 As we saw in Lecture, the general solution to the ODE is x + = C. (B now ou should be able to quickl visualize how we got that. Now using the data point (8, 6 we get (8 + ( 6 = C 64 + 36 = C C = 00. Thus the circle x + = 00 solves the IVP. Or does it? To be sure, the circle does not give us = (x (i.e., does not give as a function of x. If we want as a function of x, we have to solve for it. Since = 00 x, we have = ± 00 x, i.e., = 00 x or = 00 x. So which is it? The first is the upper semicircle, and the second is the lower. The data point (8, 6 is on the lower semicircle, so if we want a function (and not just a curve solving the IVP, we have to conclude the solution to be = 00 x. (4 On some occasions the whole curve is desired, but where we can find an actual function that is almost alwas preferable. Sometimes we need more than one piece of data. Since we ma have more than one parameter in the general solution, from algebra we know that two equations are usuall required to find two unknowns, and it is often the case here too. Next we have just such an example. Example 3 Solve the IVP = 4 (π/8 = (π/8 = 6. Later in the course we will see how to solve the ODE, and find that the general solution is = Asin x+b cosx. (You should be able to see readil that = sin x and = cosx are both solutions of the ODE, i.e., of the top line of (5, and to be able to show, if asked, that = Asin x + B cosx is a solution if not the solution of the ODE. Now we enter the data: 8 4 4 = Asinx + B cosx, = Acosx B sin x, 8 = Asin = 6 Acos This gives us a sstem of two equations in two unknowns, A and B: A + B = A B = 6 (simplifing each + B cos = = 6. 4 4 B sin A + B = 4 A B = 6. Adding the equations gives A = 0, or A = 5. Subtracting the equations gives B =, or B =. Thus the solution to the IVP is = 5 sinx cosx. (6 (5
Separable Equations Defined; Constant Solutions A first-order ODE in which the, and x, terms can be algebraicall reorganized to opposite sides of the equal sign is called separable. We have encountered and solved some of these alrea. A variation of Farlow s definition (Farlow, page 38 is of the same spirit, namel that we can algebraicall rewrite the equation into the form φ( = ψ(x, so that the -terms can be written with the differential, and the x-terms with. For our purposes, a first-order differential equation of the form = g(xh( (7 is said to be separable or to have separable variables. Note that if h(k = 0, then = k is a (somewhat trivial solution of / = g(xh(, since for the function = k, (7 gives 0 = 0 (wh?. If we take this definition, we see we can get = g(x h( h( = g(x. If we can find formulas for the antiderivatives above, we can then write the solution P( = G(x + C, (8 where P = h and G = g. It will be a one-parameter famil of curves, and if we are luck, we can solve (8 for. When we are faced with an IVP, with the ODE part separable, we need to plug the data into (8 to find C to get a curve, and possibl solve for to get an actual function. In some sense these are conceptuall the simplest differential equations to solve. In fact, the fit nicel into an calculus course when time permits. However, these equations can certainl tax one s integration skills, and solving for ma require some careful algebra. Furthermore, those constant solutions = k ma or ma not be contained in the parametrized famil of curves, as we will see in the remaining examples. If not, these are called singular solutions, as we will discuss later. Example 4 Solve the following differential equation: = x. (9 Solution: First note that = ± are both constant solutions of (9. Indeed, for = or =, both the left-hand and right-hand sides will be zero. For other solutions we see that the right-hand side is a separable product as in definition (7, and we can solve the equation as follows: = x = x = x = x + C. 3
Note that the integral on the left calls for Calc I-tpe substitution: = u = du = du = Again, putting these together gives us = ( / du u u / / + C = u/ + C = u / + C = + C. = x + C. (0 If we now wish to solve for, that is also possible (note multipling out the square is optional: = x + C = ( x + C = 4 x4 + Cx + C = 4 x4 + Cx + C = 4 x4 Cx C + = ± 4 x4 Cx C + or = ± ( x + C. In the first line, we squared both sides. The rest should be clear. However, recall that in squaring both sides we can introduce extraneous solutions, because we lose some precision in our information about the quantities involved. 3 Whether a particular solution works will rel to some extent on C (note, for instance, that we can not have C > 0 because then we would have LHS 0 and RHS > 0 in (0 see wh?. Also notice where the parameter finds itself in the final form of this solution (twice in the first form. Thus we have to be careful not to just follow the Calculus I and II mentalit of slapping a +C at the end of ever problem. Also note that if this ODE were part of an IVP, it would surel be simpler to find C in the original form (0 of the general solution; and in the final solution, we can onl have one case from the +/ if we are to have a function, which is the whole point of solving for. Further, note that the cases = ± are not possible to obtain b clever choices of C; the are outside the one-parameter famil of curves. In such a case the are called (b Farlow, p. 4 3 A simple example of how we lose information when we square both sides is the following: x = 5 x = 5 x = 5, 5. In particular we lose some information about the signs of the quantities involved whenever we square both sides, and can be in danger of admitting extraneous solutions such as the x = 5 case above. 4
singular solutions. In other examples we will have nonsingular constant solutions. 4 Finall. note that the solution to (9 is in =, = and (0, or equivalentl, ( =, =, = ± x + C. ( The previous example had moderatel simple calculus and algebra, though both did require care. Consider next the following. Example 5 Sometimes differential equations are written without being first solved for. Indeed, sometimes the and are both written as multiplicative factors at the outset. (While this seems an unnecessar complication here, it will be important later. Consider x lnxln = 0. ( It is almost alwas instructive to examine the equation as solved for = x lnxln = x lnxln. before proceeding farther: At this point we look for constant solutions as before. Again, these are solutions of the form = k for which the RHS is zero, and since the are constant solutions the LHS will also be zero. It appears that = 0 will make the -factor on the left zero, but then the ln term is undefined, so = 0 is not a valid solution. However, when ln = 0, i.e., when =, we do have a solution to the ODE (but onl for x > 0, as the reader should check. Next we separate the and x terms and integrate: = x lnxln ln = xlnx ln = xln x. The first integral ields to substitution, with u = ln impling du = : ln = u du = ln u + C = ln ln + C. The x-integral will require integration b parts, with u = lnx du = x giving us xlnx = uv v du = x lnx x dv v = x = x = x lnx 4 x + C = 4 x ( lnx + C. 4 In mathematics, singular usuall refers to something enigmatic (but still important, where the simpler mathematical analsis breaks down in some wa. However the precise meaning changes with context. For a function, a singular point might be where we divide b zero, as in the point x = 0 in the function f(x = /x, in that case resulting in a vertical asmptote, or the point x = in the function g(x = (x /(x, resulting in a removable discontinuit. In matrix theor, a singular matrix is a square matrix A n n where A does not exist, which is also one in which det(a = 0, since A = det A [adj(a] (see an linear algebra book, where adj(a = [cof(a]t, i.e., the transpose of the matrix of cofactors of A, so we are again dividing b zero if det(a = 0. In our present context, there ma be constant solutions to an ODE which are outside the parametrized famil, and these solutions are called singular. The are different from what we get from simpl varing the parameters, such as C here; there is a definite leap to these solutions, whereas man of the curves in the parametrized families can be morphed from one to another b continuousl varing C through some interval. We will see more of this phenomenon later. 5
Putting these together gives us ln ln + C = 4 x ( lnx + C, or (along with = : =, ln ln = 4 x ( lnx + C 3. (3 Again we seemed to have solved the ODE (no more derivatives present!, if we include = which is undefined in the second formula in (3, but for the second formula more can be done to solve for itself. We can take the natural exponential of both sides to then get This then gives ln = e [ 4 x (ln x +C 3] = e C 3 e [ 4 x (ln x ] = C4 e [ 4 x ( ln x ]. ln = ±C 4 e [ 4 x (ln x ], and it can t be both cases (+ and simultaneousl, so we might as well write ln = C 5 e [ 4 x (ln x ]. 4 x (ln x ] Finall, we can take the exponential of both sides again to get = e Ce[, or [ ( ] = exp C exp 4 x ( lnx. (4 Once again we see that the calculus was moderatel difficult, but then even the algebra became rather length (though no particular part of that was difficult. We reall should check to see if all constants C here work (since C = exp(±c 4 0, but we will see that if we allow C = 0 we get that constant case =, so that solution is ultimatel nonsingular. Note that though the C = 0 case technicall disappears as we move from the second formula in (3 to (4, C = 0 would have been exactl the valid case =, so it reappears in (4 if we allow C to var more than its definition (C = ±C 4 = ±e C3 would normall allow. We can see more of this phenomenon of eas, constant solutions (which ma or ma not be in the parametrized famil again in the following example. Example 6 Consider the equation = sin x. (5 Note that = and = are constant solutions of this ODE. For an other solutions we can divide (while for these two we would be dividing b zero!, and get = sinx = sin x ln = cosx + C We can thus give the general solution as ln = cosx + C. =, =, or ln = cosx + C. (6 Now in the third solution one can solve for algebraicall, and see if these two constant solutions are contained in the new parametrized famil if we allow the (new parameter to range over more than we normall would, as happened in the previous example. The details are left as an exercise. 6
Homework -A. Find all of the (infinitel man constant solutions of the differential equation = ( 5(sin exp(x3. (Please do not tr to find the general solution for this ODE!. What if we include a factor of ln, as in How does this change the constant solutions? = ( 5(sin (lnexp(x3? 3. As in Example 5, see if the constant solutions = ± for Example 6 reappear with the new parameter s extended range if we solve for in the implicit equation 4. Solve (including all constant solutions ln = cosx + C. x 3 + ( + e x = 0. Hint: First solve for. Also be slightl clever with algebra. The hard part of the answer (so ou can check is: + = ex (x + C. (This problem comes from the classical text of Rainville and Bedient, 98 edition. 5. Solve for in the previous problem. Hint: multipl both sides b, move all terms to one side of the equation with zero on the other, and use the quadratic formula but with x plaed b here, and proper care taken to identif the a, b, c terms: a 0, a + b + c = 0 = b ± b 4ac. a 7