Improper Integrals, and Differential Equations


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1 Improper Integrls, nd Differentil Equtions October 22, Improper Integrls Previously, we discussed how integrls correspond to res. More specificlly, we sid tht for function f(x), the region creted by the grph of f, the xxis, x = nd x = b will hve n re A tht is given by A = An interesting question to think bout is the following: ssuming tht f(x) is defined for ll rel numbers, wht hppens if we let, sy, the upper it of integrtion b get lrger nd lrger? For exmple, suppose we let f(x) = x, nd = 0. Then A = 0 x dx = 2 b2. We cn choose some incresing vlues for b to determine wht hppens s b gets lrger: b A By now, you cn surely tell tht s b gets lrger, so does the re. In fct, by just choosing b lrge enough, we cn mke A s lrge s we wnt. In mthemticl nottion, we write 0 x dx = 2 b2 = +.
2 To simplify nottion, we define f(x) dx = The previous computtion should come s no surprise. After ll, if you hve n infinitely lrge shpe, its re should obviously be infinitely lrge. As obvious s it my seem, this observtion my, in fct, be flse. Consider now the function f(x) = /x 2, nd let us evlute the integrl dx = x2 x 2 dx By using the Fundmentl Theorem of Clculus, we compute dx = x 2 x b 0 = b = b. The funny thing here is tht the it t the end is equl to zero. Thus, we hve dx =. x2 This is n odd result. It sys tht we cn hve n infinitely long shpe which hs finite re! This is one of the mny counterintuitive results in mthemtics which mkes the subject interesting. This leds us to the following definitions. Definition. An improper integrl is n integrl of the form f(x) dx or To compute such n integrl, we use the definitions f(x) dx = f(x) dx = 2 f(x) dx f(x) dx
3 If the its exist nd re finite, we sy the integrl converges. Otherwise, we sy the integrl diverges. Exmple. Determine if the integrl converges or diverges. 0 e x dx Solution: We strt by pplying the definition: 0 e x dx = 0 e x dx Next, we pply the Fundmentl Theorem of Clculus: 0 e x dx = 0 ex = e0 e = e Finlly, we recll the fct (which cn be obtined by exmining the grph of e x ): e = 0. So we obtin 0 e x dx =. Since this is finite number, the integrl converges, nd we sy tht it converges to. Exmple 2. Determine if the integrl converges or diverges. x dx 3
4 Solution: We proceed s before, by first writing the definition: x dx = x dx Applying the Fundmentl Theorem of Clculus, we get x dx = ln b ln = ln b. Reclling the fct (which you cn lso get by looking t the grph of ln x) we thus get tht So the integrl diverges. ln b = +, x dx = Differentil Equtions A concept which hs been of crucil importnce in modern physics is tht of differentil eqution. Roughly speking, differentil eqution is simply n eqution involving n unknown function y nd ny combintion of its derivtives. The gol is to then find function y = f(x) which stisfies the eqution. Exmple 3. A clssic exmple of differentil eqution is Newton s Second Lw of Motion, which sttes tht the force pplied to n object is equl to the product of its mss nd ccelertion. Reclling tht the ccelertion is the second derivtive of position s, this gives us F = ms. Consider now spring with one end ttched to fixed object. If we let s be the distnce of the other end from its rest position, then Hooke s Lw tells us tht the force on the spring is given by F = ks, where k is n pproprite constnt. Substituting this into Newton s Second Lw gives us ks = ms. This eqution ppers in mny brnches of science, prticulrly physics, chemistry, nd engineering. It hs solutions which look like s(t) = A cos(ωt) + B sin(ωt), 4
5 where A nd B re just constnts, nd k ω = m is clled the frequency. If we recll tht the trigonometric functions sin t nd cos t oscillte bck nd forth repetedly between  nd s t increses, we cn see why ny system which obeys this eqution is clled hrmonic oscilltor. While differentil equtions re n extremely importnt tool for modern science, it is very nnoying subject in tht there is not one method of solving them which works for ll types of equtions. In fct, for most equtions, there is no known method of solving them (though we often know tht solutions exist somehow). In ddition, the methods for solving equtions cn be quite complicted. Becuse of this, we will focus our ttention on only the two most simple types of differentil equtions which cn be solved explicitly, explined in the following definition. Definition 2. Let y be n unknown function. A Type I differentil eqution is n eqution of the form dx = f(x), where f(x) is given function. A Type II differentil eqution is n eqution of the form dx = f(x)g(y), where f(x) nd g(y) re given functions. Solving Type I Equtions To solve Type I equtions, we begin with their defining eqution: dx = f(x) Integrting both sides gives dx dx = 5
6 On the left side, we recll tht dx =. Thus dx = If we integrte the left side with respect to y, we simply obtin y + C. However, since the integrl on the righthnd side will lso dd nother constnt C, we simply combine the two constnts. Thus, we my simply write y = Exmple 4. Solve the differentil eqution dx = 2x. Solution: Bsed on the previous computtion, y is given by y = 2x dx = 2 x2 + C. Exmple 5. Solve the differentil eqution dx = cos(x)esin(x) Solution: Bsed on the previous computtion, y is given by y = cos(x)e sin(x) dx = e sin(x) + C. Solving Type II Equtions As in the cse of Type I equtions, we try to solve these equtions by simply integrting. Recll tht Type II equtions look like dx = f(x)g(y). The difficulty is tht now the right hnd side my contin expressions involving y. We try to get round this by removing nything with y by rewriting it s g(y) dx = f(x). 6
7 We then try integrting both sides: g(y) dx dx = The right hnd side looks like something we cn hndle. The left hnd side, however, looks little scry. However, we cn mke it look little nicer if we use differentils. Recll tht Then we cn rewrite Thus dx dx =. g(y) dx dx = g(y) = g(y). It my look like we hven t improved the left side t ll, but depending on wht g(y) looks like, the integrl my be quite simple. Suppose tht = G(y) g(y) for some function G(y). Then we cn t lest find n implicit solution in the form G(y) = If this is something you cn solve for y explicitly, then we hve found our solution. Exmple 6. Solve the differentil eqution dx = xy. Solution: We strt by moving the y to the left side: y dx = x 7
8 Next, we integrte y dx dx = We rewrite the left side s y dx dx = y x dx = ln y + C, nd so we get ln y = x dx = 2 x2 + C. Thus we cn get t lest n implicit solution to the eqution. We cn, however, go little further by doing the following: Suppose we wnt to cncel the ln from the left side. We cn use the following trick: recll tht the functions e nd ln cncel ech other out. Thus e ln y = y. This tells us tht y = e 2 x2 +C = Ce 2 x2. To get rid of the bsolute vlue, we note tht e 2 x2 is lwys positive. Since y is product of e 2 x2 nd C, we note tht it only mtters whether C is positive or negtive. If C > 0, then y = y, nd we re done. If C < 0, then y = y = Ce 2 x2. But we cn then move the negtive to the C: y = Ce 2 x2, nd hide it by renming C to C (using the trick of using the sme nme for everything). Thus y = Ce 2 x2. Initil Vlue Problems In mny pplictions of differentil equtions, we need to find solution to the eqution which stisfies some condition. Frequently, these conditions consist of knowing the vlue of the unknown function y t specified xvlue. Such types of problems re known s initil vlue problems, becuse we re 8
9 given y = y 0 for some x = x 0, which often represents the time t which the system strts. To solve such problems, we proceed s before, pplying the initil condition y(x 0 ) = y 0 t the very end. This will force the constnt C to be specific number, s only one solution will be ble to stisfy the initil condition. Becuse of this, we refer to solutions of initil vlue problems s specific solutions. Exmple 7. Solve the initil vlue problem dx = y y(0) = 2. Solution: We proceed s before, seprting x s nd y s: y dx = We then integrte, which (fter pplying the definition of differentils), gives us = dx. y Evluting the integrl gives ln y = x + C Solving for y in the sme wy s in the previous exmple gives us y = Ce x To continue, we must be very creful here. We know tht y(0) = 2. You my be tempted to just sy 2 = Ce 0 = C, nd so you my think tht C = 2. However, the function y(x) = 2e x does not stisfy the condition y(0) = 2. As we sid before, when your initil condition is negtive, the constnt C must be negtive. Thus C is relly C = 2. Thus, the specific solution is y(x) = 2e x. 9
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