MAT 127: Calculus C, Spring 2017 Solutions to Problem Set 2

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1 MAT 7: Calculus C, Spring 07 Solutions to Problem Set Section 7., Problems -6 (webassign, pts) Match the differential equation with its direction field (labeled I-IV on p06 in the book). Give reasons for our answer. () =, (4) = ( ), () = +, (6) = sinsin. The slopes in () are independent of, so do not change under horizontal shifts of the picture. This is the case onl in III. The slopes in (4) are 0 for = 0 (the -ais) and =, and so are represented b horizontal line segments on the -ais and the line =. This is the case onl in I. The slope in () at the origin (0,0) is negative, represented b a downward-sloping segment. This is the case onl in IV. This leaves onl II for (6). As a check, note that the slopes of in (6) vanish for = 0 and = 0 (the two aes). This is the case onl in II. () III; (4) I; () IV; (6) II Section 7., Problem (pts) Sketch the direction field for the differential equation =, = (), and then sketch a solution curve through (0, ). The direction field can be obtained b computing the slopes = = ( ) at a number of points ( i, j ) and depicting these in the -plane b short line segments through each point ( i, j ) of the corresponding slope. The main features in this case are the following. The slopes are 0 on the -ais (=0) and on the line =; these slopes are represented b horizontal line segments. These two lines, the -ais and the line =, break the -plane into 4 segments. The slopes are positive in the top region (where >0 and >, being above the line =) and in the bottom region (where <0 and <). The slopes are negative in the right region (where >0 and <) and in the left region (where <0 and >). the magnitude of the slopes increases awa (roughl speaking) from the two lines of 0-slopes. As (0,) lies on the -ais (line of 0-slopes), the slope is 0 and so the tangent line to the solution curve is horizontal. Just to the left of (0,), the slopes are negative, and so the solution curve descends to (0,) from the left. As we trace it backwards (to the left), it ascends and keeps on ascending, since it stas in the left region (where the slopes are negative and we trace them backwards). In fact, as we move to the left becomes more negative, while becomes more positive (because increases as we move to the left), and so =( ) becomes more negative. So the solution curve ascends steeper and steeper as we move to the left (following it backwards).

2 Just to the right of (0,), the slopes are positive, and so the solution curve ascends from (0,) to the right. It continues to do so until it crosses the line = (line of 0-slopes); at this crossing the curve must be horizontal. To the right of the crossing, the curve enters the right region, where the slopes are negative. It thus begins to descend. The rate of descent increases (the curve becomes steeper), as increases, because =( ) becomes more negative ( becomes more positive, while moves further down from the line =). upward slopes increasing awa from =0 and = g downward slopes increasing awa from =0 and = downward slopes increasing awa from =0 and = =f Figure : Graph of = () = f() for Problem The previous paragraph assumes that the curve actuall crosses the line =. Is this reall the case? Can t the curve just keep on going up in the top segment? In order to see that the curve should cross the line =, ou need to draw the slopes at sufficientl small intervals to the right of (0, ). However convincing the resulting picture ma be, this is not a proof (full justification), which is given below (and not required to receive full credit for this problem). Bonus 0pts (all or nothing). Let f =f() be the solution of =( ) with f(0)= and g()=+. We want to show that the graph of f crosses the line = for some >0. Since the graph of g crosses this line at (,)=(,), it is sufficient to show that g()>f() for all >0 (the graph of f then lies below the graph of g and thus must also cross the line =). Let h()=g() f(). Then h(0) = 0, h () = g () f () = ( f() ). Thus, h (0)= > 0 and so h()>0 for all >0 small (since h(0)=0 and h is increasing near 0). If >0 is the smallest number such that h( )=0 (so f has climbed back to g), then h (c)=0 for some c (0, ) b the Mean Value Theorem (see p7). For this value of c, h(c)=g(c) f(c)>0 and so 0 = h (c) = c( f(c) c ) > c( g(c) c ) = c( +c/ c ) = (c ) 0. This is a contradiction, because it sas 0 > 0. So there is no > 0 such that h( ) = 0, i.e. g()=+ > f() for all >0.

3 Note: Using Problem B, one can find that the solution to the above initial-value problem is given b () = e / e / 0 u e u / du. Once given this solution, ou can verif that it is indeed a solution (please do so!). Section 7., Problem 8 (pts) Make a rough sketch of the direction field for the autonomous differential equation =f(), where the graph of f is as shown in the first diagram in Figure. How does the limit behavior of solutions depend on the initial value of (0)? f() Figure : Graph of f() for Problem 8 and direction field for = f() Since the slope = f() does not depend on, the direction field diagram does not change under horizontal shifts. Since the graph of f is smmetric about the f-ais, i.e. f( )=f(), the slopes do not change under the reflection about the -ais, i.e. the are the same at (,) and (, ). Since f() = 0 for = ±,±, the slopes are horizontal on the four lines = ±,±, giving rise to four constant solutions. Since = f() > 0 for (, ) (,) (, ), the direction fields in these regions are upward sloping and the solution curves ascend. In the region (0) (, ), the slopes become steeper lower down and flatten near the line = ; thus, the graphs = () descend rapidl and approach = as. In the region (0) (,), the slopes flatten near the lines = ±, while peaking on the line = 0; thus, the graphs approach = as and = as. In the region (0) (, ), the slopes become steeper higher up and flatten near the line = ; thus, the graphs ascend rapidl as and approach = as. Since = f() < 0 for (, ) (,), the direction fields in these regions are downward sloping and the solution curves descend. In the region (0) (, ), the slopes flatten near the lines = and = ; thus, the graphs approach the lines = as and = as. In the region (0) (,), the slopes flatten near the lines = and = ; thus, the graphs approach the lines = as and = as. The above conclusions are restated in Figure. The first diagram shows the -ais and indicates the constant solutions b large dots. The arrows between the dots indicate whether the solutions in each interval go up and down. The second diagram shows a number of solution curves. The arrows in each region created b the constant solutions indicate whether the graphs of the other

4 solutions in each segment rise or fall; these graphs must approach the two nearb constant solutions as ±. The first diagram makes sense onl for autonomous first-order differential equations, i.e. equations of the form = f(), where is a function of some variable (note that there is no eplicit dependence on in the equation). The diagram of graphs is smmetric about the origin if the graph of f() is smmetric about the -ais (i.e. is an even function, f( ) = f()), as it seems to be. Figure does not indicate direction fields; onl solution curves are shown. The direction fields are tangent to these curves. 0 Figure : Diagrams of solution curves for Problem 8 Section 7., Problem (pts) Use Euler s method with step size. to compute approimate -values,,, and 4 of the solution of the initial-value problem =, = (), () = 0. These four values i are approimations for ( i ) with i = 0 + hi, where h =.. The are obtained inductivel b computing the slope s i = ( i, i ) at ( i, i ) and moving along the slope line from i to i+ = i +h so that i+ = i +s i h. The results of this computation are: i i i s i = i i i+ = i + s i = 0 = = 4 = 4 = 7 7 = = 4 = 49 4 The first column consists of the numbers i running from 0 to n, where n is the number of steps (4 in this case). The second column starts with the initial value of ( in this case) with subsequent entries in the column obtained b adding the step size h ( in this case); it ends just before the final value of would have been entered ( = + in this case). Thus, the first two columns can be 4

5 filled in quickl at the start. The first entr in the third column is the initial -value (0 in the case). After this, one computes the first entries in the remaining two columns and copies the first entr in the last column to the third column in the net line. The process then repeats across the second rowandsoonon. Theestimateforthefinalvalueof isthelastentrinthetable( 49 4 inthiscase). =, =, =, 4 = 49 4 Note : While the table above provides a quick wa of implementing Euler s method, it is important to understand what Euler s method does geometricall. Figure 4 shows the graph of the solution to the initial-value problem; it must pass through (,0) due to the initial condition (0) =. Figure 4 also shows the path corresponding to Euler s method with 4 steps going from 0 = to f =. Instead of following the solution curve (which we do not know), it follows the slope line from ( 0, 0 ) = (,0) until the -value becomes = /; we know the slope s 0 = 0 0 of this line and so can find the -value corresponding to the -value of = / on this slope line. Once we arrive at (, ) = (/, ), we net move along the slope line through (, ) until the -value hits = ; its slope is s =. Once the -value corresponding to the -value of = on this slope line is determined, this procedure is repeated twice more. Each of the four short directed line segments in the diagram corresponds to a row in the table. 0 0 / / ( 0, 0 ) (, ) (, ) ( 4, 4 ) Figure 4: Graph of solution to IVP in Problem and its approimation b 4-step Euler s method Note : Our estimate 4 is an over-estimate for () because <0 for all >0 and for all solutions =() used to estimate (), and so the tangent lines lie above the graphs. In order to see this, note that = < 0 if <0 and 0. Thus, a solution =() of the differential equation is decreasing for all 0 0 if ( 0 )<0. For such a solution, as claimed above. = ( ) = = < 0 0 < 0 for all 0, Note : The actual solution to this initial-value problem is () = + 4e, which can be obtained from Problem B and verified directl (please do so!). So () = 8 4e.6, while 49/4 =.. So the estimate is not so good, which happened because the solution decreases rapidl and the step size h was rather large. Note 4: There will be an Euler s method problem on the first midterm (and possibl on the final). No calculators will be allowed. You ll need to carr out the computations using simple fractions

6 (e.g. ) and simplif our final answer as much as possible (e.g = 8 ). Section 7., Problem (webassign, pts) Use Euler s method with step size. to estimate (.), where () is the solution of the initial-value problem = +, = (), (0) =. Since the initial-value of is 0 and the final is., while the step size is h=., there are n = final initial h =. 0. steps to make. Thus, we need to find estimates,,, 4, for ( i ) with = i = initial +hi = 0+ 0 i = i 0. The are obtained inductivel b computing the slope s i = ( i, i ) at ( i, i ) and moving along the slope line from i to i+ = i +h so that i+ = i +s i h. The results of this computation are: i i i s i = i (+ i ) i+ = i + 0 s i 0 0 (+0) = + 0 = ( ) = = ( ) + = ,000 = ( ) =, 6, , 6,000 = 96,8 6, ,8 6,000 96,8 6,000 ( ) + = 6,760,677,,000 96,8 6, ,760,677,0,000 =,0,7,0,000 See the previous problem for comments. Note : Our estimate is an under-estimate for (.) because <0 for all (0,.) and for all solutions =() used to estimate (.), and so the tangent lines lie below the graphs. In order to see this, note that =(+) > 0 if >0 and 0. Thus, a solution =() of the differential equation is increasing for all 0 0 if ( 0 )>0. For such a solution, = ( +) = + + = (+)+ +(+) = (++ ) > 0 for all 0, as claimed above. Note : Since the above initial-value problem involves a separable differential equation, it can be solved (please do so!) to find that the actual solution is () = e +. 6

7 Once this function is given, one can check directl that it solves the initial-value problem (please do this also!). The above formula for the solution gives So the above estimate (.) = e + 8 = e ,0,7,0,000 = is not great despite of the relativel small step size; this is because the second derivative is large. Problem B (0pts) The Fundamental Theorem of Calculus provides a quick wa of finding the general solution to differential equations of the form It turns out that ever equation of the form = f(), = (). () +a() = f(), = (), () can be reduced to (). Simpl multipl both sides of () b a nonzero function h=h() such that h = ah: h() +a()h() = h()f() h +h = hf (h) = hf. We can integrate both sides of the last equation and then divide b h. (a; pts) Show that for an function a = a(), there eists a nonzero function h = h() such that h = ah. Let f() = c a(u)du, h() = e f(), where c is an fied constant (e.g. 0). B HW Problem A, part (i), h = ah. (b; 8pts) Find the general solution of the differential equation + = e, = (). What is the relation with the solution in Section 7., Eercise #? In this case, a()= for all and so we can take h() = e 0 du = e. B the product rule, our differential equation is equivalent to ( e ) = e e = e () = e +C = () = e +Ce 7

8 The specific solution in Section 7., Eercise # is the C= case of the general solution. (c; 0pts) Find the solution to the initial value problem = +, = (), (0) =. What is the relation with the numbers in the last column of the table at the bottom of p04? Move to LHS and multipl both sides b e : e e = e ( = e ) = e = e () = e d = = e e +C = () = +Ce. ( de = e ) e d This is the general solution of the differential equation. The initial condition gives = 0+Ce 0 = C =. So the solution to the above initial-value problem is () = +e In particular, () = +e The numbers in the last column of the table are approimations to () obtained b Euler s method with progressivel smaller steps h and approach the actual value of (). Remark: One of the aims of this problem is to introduce another collection of first-order differential equation that can be solved eplicitl. One such a collection, consisting of separable equations = f()g(), is introduced in 7.. The collection (), introduced is this problem is called linear first-order differential equations. While the two collections are ver different, the differential equations () are both linear (take a() = 0 in ()) and separable (take g() = ). So are the linear equations () with f() = 0 as well as with a() and f() both constant (the latter ones are also autonomous, with no eplicit dependence on ). All of these equations can thus be solved in two different was. 8