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# MATH PHYSICAL SCI MATH 121B

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This 5 page Class Notes was uploaded by Kavon Feest on Thursday October 22, 2015. The Class Notes belongs to MATH 121B at University of California - Berkeley taught by J. Metcalfe in Fall. Since its upload, it has received 47 views. For similar materials see /class/226585/math-121b-university-of-california-berkeley in Mathematics (M) at University of California - Berkeley.

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Date Created: 10/22/15

Math 302 Differential Equations Metcalfe Summer 2001 June 18 2001 Method of Undetermined Coef cients Section 36 43 When Use this technique to solve linear nonhomogeneous equations when the forcing term consists of combinations of polynomials Sines cosines and exp s What This technique allows us to nd the particular solution for our homogeneous equation It basically consists of making an educated guess and then checking that guess How The instructions will be given for the n 2 case This method generalizes unchanged to higher order equations though 0 Write the equation in the form 24 WW qty W The general solution will be of the form W 24005 Y0 where go is the solution to the corresponding homogeneous equation and Y is any solution to the nonhomogeneous equation 0 We rst need to nd the complimentary solution ye lt solves the homogeneous equation 24 WW 6105 0 This can be done using the methods of Sections 31 34 35 42 using series or using Euler s equations 0 We need to guess the form of Y based on the form of g 1 lfg Pn antnan1t 1a1ta0 then let Yt AntnAn1t 1A1tA0 Your guess for Y must be the general nth order polynomial So even ifg t2 you need Yt At2 Bt C not just the t2 term lfg 5 then let Y Aeat lfg sin t or g cos t then let Y Asin t Bcos t If g 91 92 then nd Y1 corresponding to 91 and Y2 corresponding to 92 separately and let Y Y1 Y2 rPPJEO 01 If g 91 92 then let Y Y1 Y2 where Y1 is the guess corresponding to 91 and Y2 is the guess corresponding to 92 0 Now we must examine ye If any piece of our guess for Y already appears in ya we must place t s in front of it until it is longer in yo 0 Take the necessary derivatives of Y and plug into the original differential equation o Solve for the constants in Y 0 Form the general solution y yo Y o If given use an initial conditions to resolve for the unknown constants 0 Check your solution by taking the derivatives and comparing it to the original equation Examples 1 Find a general solution of y 7 y 10541 c The coefficient of y is already 1 so there is nothing to do here 0 Find ye lt solves y i y 0 We know from the previous sections that ye Clet 0264 0 Here gt 105 so our guess should be Yt A5 4 0 Since 5 t is not in our complimentary solution there is nothing to be done here 0 Y t 4A5 and Y t 1614 so plugging in gives us 16Ae4t 7 A5 151454t 1054 0 Solving for A yields A g 0 Thus the general solution is given by 2 yt clet 3267i 354 2 Find a general solution of 7 t y 7 y 7 105 o The coefficient of y is already 1 so there is nothing to do here 0 Find ye lt solves y i y 0 We know from the previous sections that ye clet 0254 0 Here gt 10 so our guess should be Yt Act 0 Since at is a member of our complimentary solution we need to multiply by 25 Thus Yt Atet Since 255 is not a member of the complimentary solution we may continue with this guess o Y t Act Atet and Y t Atet 2Aet Plugging in gives us Atet 2Aet 7 Atet 2Aet 105 0 Solving for A gives A 5 0 Thus the general solution is given by yt clet l 3267i l Stet 3 Find a general solution of y 37 2y 2t740cos2t o The coefficient of y is already 1 so there is nothing to do here 0 Find ye lt solves y y 7 2y 0 We know from the previous sections that yo Clet 62572 Here 91t 225 so our guess should be 3605 At B 0 Since there is no term of our complimentary solution of the form clt or 01 we may continue without change 0 Y t A and Y t 0 So plugging in gives A72At B 72At A72B 2t 0 Solving for A and B gives A 71 and B 7 Thus 1 Y1t Here 92t 740 cos 225 so our guess should be Y2 Asin 2t B cos 2t 0 Since there is no sin 2t or cos 2t in our complimentary solution we may continue without change 0 Y2 2A cos 2t 7 2B sin 2t and Y2 74A sin 2t 7 4B cos 225 Thus plugging in gives 74A sin 2t74B cos 2t2A cos 2t72B sin 2t72A sin 2tB cos 2t 76A72B sin 2t76B2A cos Solving for A and B gives 76A 7 2B 0 and 76B 2A 740 Thus A 72 and B 6 Therefore Y2 72sin2t6cos2t 0 Thus the general solution is given by 1 Mt Clet 62672 7 90 7 5 7 2 sin 2t 6cos 2t 4 Find a general solution of y y at cost The leading coefficient is already 1 so there is nothing to do here Find ye lt solves y y 0 We know from the previous sections that yo C1 6275 6354 Here 975 91 92 6t cost Thus our guess should be Wt Y1 Y2 it Asint Bcost Notice that we don t need a C for the guess corresponding to 5 since each term already has a constant Since neither 5 cost nor 5 sint are members of our complimentary solution we may continue without alteration Y t A 7 Betsint A Bet cost Y t 72Betsint 2145 cost and Y t 72B 7 2Aet sint 2A 7 2Bet cost Thus plugging in gives 72B 7 2B 7 2Aet sint 2A 2A 7 2B cost at cost Solving for A and B 74B 7 2A 0 and 4A7 2B 1 Thus A and B 7 Thus the general solution is given by 1 1 2475 C1 Cgt 6357 get sint 7 Est cost Here is one for you to try 5 Find a general solution of y 7 229 7 3y 6t62t 0 Check that this is a linear nonhomogeneous equation Divide through by the leading coe icient if necessary 0 Find the complimentary solution yo 0 Make a guess at Yt o ls any piece of Y already included in go If so make the necessary alterations 0 Take derivatives of Y and plug in Solve for the constants 0 Form the general solution 0 Check your solution

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