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Spring 2004 - Bender's Class - Exam 2

Spring 2004 - Bender's Class - Exam 2 - Math 20D Second...

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Unformatted text preview: Math 20D Second Midterm 48 points May 26, 2004 0 Print Name> ID number and Section on your blue book. 0 BOOKS and CALCULATORS are NOT allowed. One side of one page of NOTES is allowed, 0 You must show your work to receive credit. Some integrals and derivatives that may be useful: ftancr dm = iln(cos Cc) + C (tancr)’ = sec2 :r (seem)’ = sec a tanm fsecm dm : ln(sec m + tan m) + C (arctan m)’ : #5 (arcsinm)’ : 1; 1. (27 points) Find the general solutions of the following differential equations (a) @563”)? :1 (b) fizz/+9 : 44 (C) (m2+1)9’*y : 0 1. (a) An integrating factor is exp (f —tanm dz) : cos 1. Thus (3/ cos a)’ : cosm and 1 ' C 9(1) : fcosmdr : $ : tanm+0secm cosm cosm (13) Since the characteristic equation for the homogeneous equation is 0 : r2—2r+1 : (r—IY, y : Gist + Ogtet is the general solution to the homogeneous equation. By un- determined coefficients, a particular solution is y = At + B. Since 3;” = A and y” = 0) we have 4: = 072A+CAt+Bj = At+CB72Al Thus A : 4 and B : 8, The general solution to (b) is therefore y : Clef + my + 47: +8 (c) Rearrange, separate variables and integrate: is a dm : /1+m2 and so lny : arctanm+O You may leave the answer this way, with or without absolute values on y, or you may solve for y. 2. (3 points) I have decided to find a series solution 3/ : Z anm” to (4 + mg” + (1 7 my : 0. For what values of m can you guarantee that the series will converge? Why? You must give a reason—the ‘Why?”—to receive credit. 2, Since 4 + 032 = U for o: = :21, the radius of convergence of the series for 41%;; is 2, Thus the best we can guarantee is lml < 2. 3. (9 points) 3/ = t and y = t2 are solutions to $2912, 2th + 23/ = 0. Find a particular solution to tgy'” 7 2:3/ + 23} : tzef. You may leave integrals in your answer, 2 3, Since 71’ e 2y“/t+y/t2 = 8* and WW2] = li ta: $28t 2 tat s 2 start y:—tft—2dt+tft—2dt:—t/edt+tft 4, (9 points) Mt) satisfies the differential equation = t2, a particular solution is 2/(i)+y(t-1):1, with y(t) : U for t S 0, Compute the Laplace transform Y(s) : £{y(t)}. 4, By the table for Laplace transforms, 1 sY(s) +y[0)+e’5Y(sj = 1/5 and so Y(s) = ml When your exam is returned, it will have the grade you will receive if you do NOT talre the final exam. In computing that grade) this exam will be weighted more than the first exam to reflect the fact that the final will emphasize differential equations. END OF EXAM ...
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