soln5_082

# soln5_082 - MATH2010/2100 Assignment 5 V and Tutorial Set 5...

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Unformatted text preview: MATH2010/2100 Assignment 5 V - and Tutorial Set 5- K6.3, p.240: Questions 4,-8.*, 14, 24, 28. s 506.5, p.253: QueStions if, 18, _ K6.7, p.263: Questions 12, 3*, 14. Solutions to the three starred problems to be handed in at the end of your tutorial or placed in your assignment box on Level 3, Building 67 , by 5pm Monday, September 1. NB: Use a cover sheet! CHAP. 6 Laplace Transforms G PROJECT. Shifting Theorem. Explain ‘ les of the two shifting ations and examples. 24:3 nan“? STE? FUN "‘ SWWENG ’E‘HELQREM Sketch or graph the given function (which is assumed to be zero outside the given interval). Represent it using unit step functions. Find its transform. Show the details of your work. 2. t (0 < r < l) sin3t(0<t< 77) 3. e‘(0<z<2) 5.t2(1<t<2) 7.cos1rt(l<t<4) (5<:< 10) t(3<t<6) 6. :2 (r > 3) ‘&1.~e‘t(0<t<n)9.t - 10. sm an (t > 61r/w) 11. 20 cos 7r 12. sinht(0<t<2) 13. e"t (2<t<4) tNQﬁRSE TRANSE‘ﬁRMﬁ EW "i'ﬁii SECQNE} ﬁﬁlﬂii‘éﬁ FMFOREM Find and sketch or graph f0) if EEG) equals: 14. se‘s/(s2 + wz) - 15. e"‘1s/s2 2 " (5—2 + 5"1)e"s 16. s" 17. (e’ztrs ~ e‘8"S)/(sz + 1) 18. e‘7T3/(s2 + 25 + 2) 19. e 20. (l - e's+k)/(s - k) 21. se’ss/(s2 22. 2.5(e‘3'85 ~ e'Z-“ws (\$332.3. ,._,._.o<. ~25/S5 ERE‘E‘EM. VAtﬁE aaeeems, ERQME Wifti ﬁt\$€€§i~éﬁt€§§0i§\$ lhéﬁﬁﬁ lace transform and showing y(0) = O. 2.33% r...” Using the Lap the details, solve: 23. y" + 2y' + 2y = 0, Vm=1 24- 9y" __ 6y: + y "ﬂm=1 25. y" + 4y' + 13y = rm=m = 0, y(0) = 3, M) = l—‘é. Mﬂlﬁlﬁ 0‘? at. 35. (Discharge) 30. y" — 16y = Oift>4', 31. y" + y’ 0 < t <I27r and3 sin2t y(0) = 1. 32.y"+8y 0<t<2and0ift>2', ﬂm=4 33. (Shifted. data) y” +‘ 4y if r > 5-, ya) = 34. y'rrvl- Zy' + 5y — - 2y = r(t), r0) = 3 sint ~ cost if — cos 2t if t > 2n; ym=0 ’ + 15y = r(t), rm = =812if0<t<5and0 1+ cos2,y’(1)=4‘—2sin2 lOsintifO<t<2rrand0if = 26”"- 2 t > 211", y(qr) = l, y'(7r) EC'E'REEL CEMLEEE'E'S Using the Laplace tr apacitor of capacit harged so that its p 0. ansform. find the charge q(t) on the c ance C in Fig. 125 if the capacitor is c otential is V0 and the switch is closed at t = Fig. 115. Problem 35 Rﬁ-QERCUW place transt circu the current at I 36—38 Using the La the current i (t) in the C = 10—2 F, where rm and showing the details. ( it in Fig. 126 with R = 10 f/ = O is assumedt ’ zero, and: 36. 120) = 100 V if 0.5 < t < Why does i(t) have jumps? ‘ 37.0 = 0ift<2and100(t—- 2)Vift?2 38. v = 0 ift< 4and14-106e‘3‘v ift>. v(t) Fig. 126. Problems 36— 0.6 and 0 mile 38., {gm} eminent? / Using the Laplace transform ircuit in Fig. the current Kt) in the c and: and showing the A 127 , assumi. m: Find by integration: 1. 1 a 1 3. t =I< et 5. l t cos wt 7' en: ,1: e—kt {gig iNVERSE tamsroms av commotion Find f(t) if 580‘) equals: 9. 1 l 11. —-—— s(s2 + 4) (s - 3)(s + 5) ral web“ ,A Volterra integral Equation of the Second Kind ' Soi'veithe Volterra integral equation of the second kind3 t yir) “ J-WT) Sin (1 — 70517 = r, 0 \ Solution. From (1) we see that the given equation can be written as a convolution, y — y t sin I = t. Writing Y“: 513(3)) and applying the convolution theorem, we obtain 1 2 s Y(s) — Y(s) 32 + 1 = Y(s) 52 + = 1 E . The solution is 52 + 1 1 1 Y8) = AA = 23 + “5—4— and gives the answer Check the result by a CAS or by Substitution and repeated integration by parts (which will need patience). Another Volterra Integral Equation of the Second Kind Solve the Volterra integral equation 1: (1+T)_V(f-‘r)d'r=l—sinht. 0 W) - Solution. By (1) we can write y — (l + t) *y = l — sinh r. Writing Y = \$01), we obtain by using the convolution theorem and then taking common denominators l 1 l 52 — s — 1 :2 — 1 - s y(5) 1 ..__ __ + -—2 = i; — hence I 2 = ' S s s 32 - 1 5(52 - l) (.92 —- s - l)/s cancels on both sides, so that solving for Y simply gives 3 as) = T— y(t) = cosh t. .1 - 1 and the solution is CONVQaﬁﬂﬁNS 3! iNTEﬁRATiGN 1 s 13' s2(s2 + 1) 2 + 16)2 1* t 1 5 eat * ebt (a 95 S(Sz _ 16. (S2 + 1X52 + 25) - 1 * f0) , sin t t: cos 2 17. (Partial fractions) Solve Probs. 9, 11, and 13 by using partial fractions. Comment on the amount of work. sowme mama, VALUE momma Using the convolution theorem, solve: 1 1.8. y” + y = sint, y(0) = o, y’(0) = 0 3(5 ‘ 1) 19. y” + 4y = sin 3:, y(0) = 0, y’(0) = 0 1 .g}_"+5’+4=2-2t (0)=0 ___ -y y y e . y , sic—2) y’(0)=o 10 12. 3If the upper limit of integration is variable, the equation is named after the Italian mathematician VITO VOLTERRA (1860—1940), and if that limit is constant, the equation is named after the Swedish mathematician IVAR FREDHOLM (1866—1927). “Of the second kind (ﬁrst kind)” indicates that y occurs (does not occur) outside of the integral. " 9Y1 + 6Y2: 3’2(0) = “3 —- 15 cost + 27 sint, yz -- 150 sin t, Yam) = 2 ‘1 + 33:2. y'z = 4y: — yz. ~ ' y2(0) = 3 rial — u(t - 1), ~ ;+ 1 — MU — 1). m0) = o, tl’l‘ Y2w y; = 4)’1 ’l' 2» + 64tu(t — 1), 2. y2(0) = 0 + euo — 2V“, y; = Y1 + 2y» 07 = 1 fyz, y; = —y1 + 2[1 — u(t — 277)] cost, a 1, = 0 43y1 + yz + u(t — l)et, V ~4y1 + 2y2 + u(t - l)et, "(0) = 0: yzlo) = 3 42% +I2y2, yfé = 2)’1 — 5y2. m0) = 1, y1(0) = 0, y2(0) = 3, MO) = 0 32’: = 4)’1 + 8.312. y'é = 5y1 + yz. do) = 8’ yi(0) = #18, MW) = 5: 32(0) = “21 is’1’+ y2 = —101 sin 10:, yg + y, = 101 sin 10:, Vim) = 0, y'1(0) = 6. mm = 8, y2(0) = ‘6 2et + e't, y; + )1; = 2 sinh t, t -yi+yé I I )‘3 + 3’1 = e . yi(0) = 0. y2(0) = 1, y3(0) = 1 , 20. 4y; + y'z — 2y; = 0, —2y’1 + y; = 1. 1 2y; — 4y; = —l6t ‘ M0) = 2, y2(0) = 0, y3(0) = 0 21. TEAM PROJECT. Comparison of Methods for i Linear Systems of ODEs. - (a) Models. Solve the models in Examples 1 and 2 of Sec. 4.1 by Laplace transforms and compare the amount of work with that in Sec. 4.1. (Show the details of your work.) (b) Homogeneous Systems. Solve the systems (8), (l 1)—(13) in Sec. 4.3 by Laplace transforms. (Show the details.) (c) Nonhomogeneous System. Solve the system (3) in Sec. 4.6 by Laplace transforms. (Show the details.) ¥URYMER AWLSCAﬂGNS 22. (Forced vibrations of two masses) Solve the model in 23. 24. 25. Example 3 with k = 4 and initial conditions 311(0) = 1, y{(0) = 1, 312(0) = l, y£(0) = —-1 under the assumption that the force 11 sin I is acting on the first body and the' force —11 sin ton the second. Graph the two curves on common axes and explain the motion physically. CAS Experiment. Effect of Initial Conditions. In Prob. 22, vary the initial conditions systematically, describe and explain the graphs physically. The great variety of curves will surprise you. Are they always periodic? Can you find empirical laws for the changes in terms of continuous changes of those conditions? (Mixing problem) What will happen in Example 1 if you double all ﬂows (in particular, an increase to 12 gal/min containing 12 1b of salt from the outside), leaving the size of the tanks and the initial conditions as before? First guess, then calculate. Can you relate the new solution to the old one? (Electrical network) Using Laplace transforms, ﬁnd the currents i1(t) and i2(t) in Fig. 146, where v(t) = 390 cost and 13(0) = O, i2(0) = 0. How soon will the currents practically reach their steady I 4 state? Network Currents Fig. 246. Electrical network and currents in Problem 25 26. (Single cosine wave) Solve Prob. 25 when the EMF (electromotive force) is acting from O to 2n only. 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soln5_082 - MATH2010/2100 Assignment 5 V and Tutorial Set 5...

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