MA Exam 3 Memo Monday, November 17, :30pm in Elliott

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1 MA Eam 3 Memo Moda, November 17, :30m i Elliott 1. Eam 1 covers sectios 6.1, 6.2, 6.3, 6.5, Z1.1, Z4.3, Z4.4, Z11.2 (Lessos 18-27) 2. The eam will cosist of 12 multile choice roblems. The eam is machie graded, so there will be o artial credit available. 3. Ol a oe-lie scieti c calculator is allowed o a eam. No two-lie or grahig calculators are allowed. 4. Sice the eam will be machie graded, the ol thig that will be graded is the scatro aswer sheet. You ma kee the eam form. Double check what ou are turig i to make sure ou have the correct sectio umber ad eam form reorted ad bubbled i. We caot grade athig o our eam form. 5. There are review roblems attached to this memo. It is highl recommeded that ou review old homework assigmets for additioal review. 6. Please reread the sectio o the sllabus regardig the eam. 7. You must use our 10 digit Purdue ID umber o the eam scatro. Please double check ad make sure that ou have lled it i correctl, as this is how it gets uloaded to Blackboard. 8. You must brig our Purdue ID to the eam. 9. If ou miss the eam, ou eed to cotact the course coordiator immediatel (orris@urdue.edu, MATH 810). Do ot wait util the et class sessio to cotact the course coordiator. 10. The eam is self-elaator. Istructors ad roctors are ot allowed to iterret a of the questios for a studet. 11. A studet that does ot have a valid, documeted reaso for missig a eam ma still be allowed to sig u for the make-u eam with a grade ealt. Carelessess i kowig the correct time ad lace of the eam will ot be a valid reaso for missig the eam. 12. Durig the eam, o studet is ermitted to leave before the rst 20 miutes of the eam. No studet is allowed to come i ad take the eam after the rst 20 miutes. If a studet shows u more tha 20 miutes late, the should seak directl with the course coordiator about the ossibilit of takig the make-u eam. If the do ot have a valid ad documetable reaso, a grade ealt will be imlemeted o the make-u eam. Toics Covered Fourier Series (Z11.2) f() = a X a cos =1 Z + b si for a fuctio de ed o the iterval ( ; ) Z where a 0 = 1 a = f()d 1 f() cos d b = 1 f() si d For a costat fuctio de ed b f() = k o ad iterval (a; b) ( ; ) ; the coe ciets of the Fourier series ca be determied b a 0 = k (b a) a = k si b si a b = k cos a cos b Z 1

2 If f() is liear i the form f() = k + m, the coe ciets of the Fourier series ca be calculated as a 0 = 1 a = k + mb b = k + ma k(b a) + m 2 b2 a 2 si b cos a k + ma k + mb si a cos b + m 2 2 cos b + m 2 2 si b cos a si a Verifig solutios of di eretial equatios (6.1) Searable di eretial equatios (6.2) Solvig for geeral ad articular solutios of di eretial equatios b searatio of variables: N()d = M()d Solvig rst-order liear di eretial equatios b itegratig factors: (6.3) d Of the form: + P () = Q() d Itegratig factor: e R P ()d Alicatios of di eretial equatios (6.3, 6.4) Growth ad deca: d dt = k Newto s Law of Coolig: Families of orthogoal trajectories Geeral rate roblems d dt = k( T m) Solvig higher order homogeeous di eretial equatios (6.5). Solvig auiliar equatios to determie solutios to higher order homogeeous di eretial equatios. (Z4.3) If m 1, m 2,...,m are solutios of the auiliar equatio, the the geeral solutio of the di eretial equatio is give b = C 1 e m1 + C 2 e m2 + ::: + C e m If a auiliar equatio has a root m = a reeated times, the the geeral solutio of the di eretial equatio is give b = C 1 e a + C 2 e a + C 3 2 e a + ::: + C 1 e a. If a auiliar equatio has a root m = abi, the the geeral solutio of the di eretial equatio is give b = e a (C 1 cos b + C 2 si b) Solvig ohomogeeous di eretial equatios First d the solutio to the homogeeous solutio, c. Use method of udetermied coe ceits to d. If f() is of the form e k Use this Ae k ::: A + B 1 + ::: 2

3 Practice Problems 0 3 < < 0 1. Give f() = 2 0 < < 3, determie a 0 ; a ; ad b of the Fourier series. 5 2 < < 0 2. Give f() = 5 0 < < < < 0 3. Give f() = 4t 0 < < 7, determie a 0 ; a ; ad b of the Fourier series., determie a 0 ; a ; ad b of the Fourier series. 4. Show that = e 4 ad = e 2 are both solutios to the di eretial equatio d2 d 2 2 d d 8 = 0 5. It was determied eerimetall that 5% of a quatit of oloium-210 decaed after 20 ears. Determie the half-life of Poloium A bacteria culture grows at a rate roortioal to the umber of bacteria reset. If the size of the culture doubles i 2 hours, how log will it take for the size to trile? 7. A bod is take out of a freezer ket at 15 o F is laed i a room whose temerature if 60 o F. After 3 miutes the temerature of the bodi has rise to 5 o F. Fid the time it takes for the temerature of the bod to rise to 50 o F. 8. The isothermal curves of a metal late are give b = C. Fid the curves alog which the heat ows (the orthogoal trajectories) 9. Fid the famil of orthogoal trajectories to 2 = C A 50 gallo tak is full of a solutio cotaiig 20 lb of salt. Startig at time t = 0, ure water is admitted to the tak at a rate of 5 gal/mi, ad the well-stirred solutio is withdraw at the same rate. (a) How log will it be before the solutio will cotai half of the origial amout of salt? (b) Calculate the ouds of water remaiig i the tak after 5 miutes. 11. Fid the geeral solutio to the followig di eretial equatios. Elimiate a logarithms i the solutios, if ossible (a) 0 + = 1 (b) d d = 0 (c) 3 d d = 4 (d) d d 5 = 2 3 (e) = 0 (f) d2 d 2 9 = 0 (g) = 0 (h) = 0 (i) 3 d2 d 2 4 d d + = Fid the articular solutio of the di eretial equatio usig the give values. (a) d = d; (1) = 2 (b) 2 d = (1 + )d; (0) = 1 3

4 (c) 0 2 = 4; (0) = 3 (d) 00 4 = 0; (0) = 0; 0 (0) = 6 (e) = 0; (0) = 2; 0 (0) = 3 (f) = 0; (0) = 4; 0 (0) = Use the method of udetermied coe ciets to d the articular solutio to the followig di eretial equatios. (a) = 4e 3 (b) = (c) = Match the followig di eretial equatios with a ossible grah of its solutio (a) = 0 (b) 00 + = 0 (c) = 0 (d) = 0 (e) = 0 I II III IV V Aswers 1. a 0 = 2 a = 0 b = 2 (1 ( 1) ) 2. a 0 = 0 a = 0 b = 10 (1 ( 1) ) 3. a 0 = 14 a = 28 (( 1) 1) b = 28 ( 1) 4. (check) ears hours miutes 8. 2 = C = C 4

5 10. (a) 6.93 miutes (b) lb 11. (a) = 1 + Ce (b) 2 3(1 ) 2 = C (c) 3 4 = C (d) = 3 + C 5 (e) = C 1 e 2 + C 2 e (f) = C 1 e 3 + C 2 e 3 (g) = e C 1 cos C 2 si (h) = e 5 4 C 1 cos 4 + C 2 si 4 (i) = C 1 e + C 2 e (a) = 2 (b) 2 3 = (c) = (d) = 2 + 5e e e2 (e) = 3 2 cos si 2 (f) = 4e (a) = 2e 3 (b) = e 2 (c) = (a) I (b) III (c) V (d) II (e) IV 5