tapia (jat4858) – HW04 – clark – (52990) 1This print-out should have 20 questions.Multiple-choice questions may continue o nthe next column or page – find all choicesbefore answering.001 10.0 pointsUse the direction field of the differentialequation y′= y5to sketch a solution curvethat passes through the point (0, 1).1.−22−22xy2.−22−22xy3.−22−22xy4.−22−22xy5.−22−22xycorrect6.−22−22xyExplanation:The vector field for y′= y5with a samplecurve through our initial condition y(0) = 1looks like the followingtapia (jat4858) – HW04 – clark – (52990) 2−22−22xy002 10.0 pointsConsider the following graph of y(x)246xy−π2π2π−πChoose the equation whose solution isgraphed and satisfies the initial conditiony(0) = 2.1. y′= −y sin(2x)2. y′= y sin(2x) correct3. y′= x sin(2y)4. y′= y sin(4x)5. y′= y cos(2x)6. y′= y sin(x)Explanation:Looking at the graph of y, we can see thaty′π2= 0 and yπ2≈ 5.4. Using this, wecan see the following choices cannot have oursolution:y′(x) = y sin(x)y′(x) = y cos(2x)y′(x) = x sin(2y).In each of these opti ons, y′π26= 0.Now we can notice that y′π4> 0 andyπ4> 0. This rules outy′(x) = −y sin(2x)y′(x) = y sin(4x).In the first one, y′π4< 0, while in the secondy′π4= 0. Hence we are left with onlyy′= y sin(2x)Now we will check this answer by generatingthe direction field for y′= y sin(2x).246xy−π2π2π−πAs we can see from the above plot, ourgraph of y is indeed a solution to the differen-tial equation y′= y sin(2x).tapia (jat4858) – HW04 – clark – (52990) 3003 10.0 pointsConsider the graph2−2−4246−2xyUsing the above graph of y( x ), choose theequation whose solution is graphed and satis-fies the initial condition y(0) = 4.1. y′= y3− x22. y′= y − x3. y′= y2− x4. y′= y − x − 15. y′= y − 1 correct6. y′= −y + 4.57. y′= y2− x2Explanation:Let us begin by approximating the tangentline about y(0).2−2−4246−2xyFrom the above graph, we can see the tan-gent line about y(0) has a slope of about 3.This tells us that y′(0) ≈ 3. Now, only threeof o ur answer choices match this property.Those a rey′1(x) = y(x) − 1y′2(x) = y(x) − x − 1y′3(x) = −y(x) + 4.5Now we’ll look similarly at x = −4. Noticethat y′(−4) ≈ 0 and y(−4) ≈ 1. If we use thisvalue in our remaining 3 answer choices, wegety′1(−4) = y(−4) − 1≈ 1 − 1= 0y′2(−4) = y(−4) + 4 − 1≈ 1 + 4 −1= 4y′3(−4) = −y(−4) + 4.5≈ −1 + 4.5= 3.5.Now we can see o nl y y′1(x) can be the deriva-tive of our function. To show that it works,tapia (jat4858) – HW04 – clark – (52990) 4let us now look at a direction field of y′1.2−2−4246−2xyFrom this, we can see that the given graph isindeed a solution to the different ial equationy′= y − 1.004 10.0 pointsUse Euler’s method wit h step size 0.4 to es-timate y(2.2), where y(x) is the solution ofthe initial-value problemdydx+ 2x2y = 9x2,y(1) = 5.1. y(2.2) ≈ 4.9 422. y(2.2) ≈ 3.3 533. y(2.2) ≈ 2.3 434. y(2.2) ≈ 1.9 345. y(2.2) ≈ 4.5 9 correctExplanation:In Eul er’s Method, we predict t he value ofyiusing yi−1, y′i−1, and dx with the followingequationyi= yi−1+ y′i−1dx.Notice that y′(x, y) = 9x2− 2x2y andy(1) = 5. Hence, in this problem y0= 5,y′0= y′(1, 5) = −1 and dx = 0.4. In order tovisualize this bett er, let us use the foll owingtable to display our valuesi x y y′0 1.0 5. 000 -1.0001 1.4 4. 600 -0.3922 1.8 4. 443 0.3683 2.2 4. 590 -0.875Notice i n this table, y is computed in eachrow using the values for y and y′in the pre-vious row while y′in each row is determinedusing y and x in t he current row. Hence,y(2.2) ≈ y3= 4.59 .005 10.0 pointsUse Euler’s method with step size 0.1 t o com-pute the approximate y-values y1, y2, and y3of the solution of the initial-value problemy′= 1 + 6x − 3y, y(1) = 2.1. y1= 2.200, y2= 1.800, y3= 2.0612. y1= 2.100, y2= 1.770, y3= 2.0413. y1= 2.100, y2= 2.370, y3= 2.5884. y1= 2.100, y2= 2.230, y3= 2.381 cor-rect5. y1= 2.200, y2= 1.880, y3= 2.364Explanation:In Eul er’s Method, we predict the value ofyiusing yi−1, y′i−1, and dx with the followingequationyi= yi−1+ y′i−1dx.Recall that y′(x, y) = 1 + 6x − 3y andy(1) = 2. Hence, in this problemy0= 2, y′0= y′(1, 2) = 1 and dx = 0. 1.In order to visualize this better, let us use thefollowing table to display our valuesi x y y′0 1.0 2. 000 1.0001 1.1 2. 100 1.3002 1.2 2. 230 1.5103 1.3 2. 381 1.657tapia (jat4858) – HW04 – clark – (52990) 5Notice in this table, y is computed i n each rowusing the values for y and y′in the previousrow while y′in each row is determined usingy and x in the current row. H ence,y1= 2.100, y2= 2.230, y3= 2.381 .006 10.0 pointsIf y0satisfies the equations4dydx+3x y3= 0, y(1) = 2,for x, y > 0, find the va lue of y0(e).1. y0(e) = 51/32. y0(e) = 1 31/4correct3. y0(e) = 1 91/44. y0(e) = 1 01/45. y0(e) = 1 11/3Explanation:The di fferential equation4dydx+3x y3= 0becomes4Zy3dy = −3Z1xdxafter separating variables and integrating.Thus the general solution of this differenti alequation is gi ven byy4= −3 ln x + C,which in explicit form can be writt en asy(x) = (C −3 ln x)1/4with C an arbitrary constant. For the par-ticular sol ut ion y0this constant is determinedby t he condition y(1) = 2, for theny(1) = 2 =⇒ 24= C.Hencey0(x) = (24− 3 ln x)1/4,in which casey0(e) = (24− 3)1/4= 131/4.007 10.0 pointsIf y = y0(x) is the solution of the differentialequationp25 − x2dydx+ 2xy = 0which satisfies y(5) = 9, find the value ofy0(0).1. y0(0) = −9e102. y0(0) = −9e−103. y0(0) = 9e10correct4. y0(0) = 9e−105. y0(0) = 9e12Explanation:The differential equat ionp25 − x2dydx+ 2xy = 0becomesZdyy= −Z2x√25 −x2dxafter separating variables and integrati ng. Toevaluate the right hand side we use the sub-stitution u2= 25 − x2. Then u du = −x dx,soln y = 2Zdu = 2p25 − x2+ C.tapia (jat4858) – HW04 – clark – (52990) 6Thus the general sol ut ion of the given differ-ential equation is given byy(x) = eCe2√25−x2= Ae2√25−x2with A an arbitrary constant. For the partic-ular solution y0, the value of A is determinedby t he condition y(5) = 9, sincey(5) = 9 =⇒ A = 9.Hencey0(x) = 9e2√25−x2,from which it follows thaty0(0) = 9e10.008 10.0 pointsIf y0satisfies the equations(x2+ 16)dydx= xy, y(0) = 4,determine the va lue of y0(20).1. y0(20) = 5 ·
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