Calculus and Analytic Geometry IV
Calculus and Analytic Geometry IV MATH 254
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This 4 page Class Notes was uploaded by Theresia Hyatt DDS on Monday September 21, 2015. The Class Notes belongs to MATH 254 at Ohio State University taught by Staff in Fall. Since its upload, it has received 46 views. For similar materials see /class/209947/math-254-ohio-state-university in Mathematics (M) at Ohio State University.
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Date Created: 09/21/15
Important Concepts for Exam 27Math 255 The following are a list of important things to know for the rst exam General Material 0 Computing derivativesiAs the name suggests ordinary derivatives are ubiquitous While one should also know the various techniques to com pute derivatives the exam problems should not require any special tech niques beyond possibly the chain rule product rule or quotient rule Furthermore the problems will avoid using functions more complicated than polynomials sines cosines exponential functions or logarithms 0 Computing antiderivativesir1de r1ite integralsiThis is an absolutely es sential skill for this course One absolutely must know how to per form integration by parts and integration by substitution also called u substitution The test problems will avoid using trigonometric sub stitution and some of the more exotic integration techniques Finding roots of polynomialsiln the future one should practice their techniques for nding roots of polynomials and factoring polynomials for linear equations with constant coef cients For exam 2 one should be comfortable with handling polynomials of degree three and possibly degree four By this point in the quarter one should feel comfortable verifying that a function is a solution to a differential equation Recall that to verify that a function Ms solves a differential equation a person must set y x take the appropriate number of derivatives x gt and substitute these into the differential equation A person then makes sure that the two sides of the equation agree For example to verify that 62m is a solution to dzy dy 7 i 47 4 0 dx2 dx y 7 we set y 62m and notice that y 262m and y 462m Thus we see that 45 i 4252w 452w 45 7 85 45 0 This veri es that 62m is a solution to the given differential equation Chapter 3 amp 4 Material Because the material in Chapter 4 simply generalizes the material in Chapter 3 to higher order differential equations we address all of the neces sary exam material in this list 0 Given one solution to a homogeneous second order differential equa tion One should be able to use reduction of order to nd the general solution to the differential equation Problems 23 30 of Section 35 contain such problems including one that was part of a homework as signment ln 23 of Section 35 the problem calls for one to solve the differential equation dzy dy tziizlti 6 0 tgt0 dtZ dt y given that y t t2 is a solution 0 For homogeneous linear differential equations with constant coef cients one should know how to interpret characteristic polynomials that have repeated roots or complex roots For instance if our charac teristic polynomial is A2 4 i 33 then then it has roots A i2 and A 3 a triple root Recall complex roots will contain sines and cosines and multiple roots are handled with the multiplication of powers of t Thus in our example the general solution to such a differential equation would be y gtt 01 cos2t Cg sin2t 0365 04125 O5t263t Also understand how to interpret solutions coming from roots of the form 04 i 2396 These have the form y gtt Clem cos t 0260quot sin t Thus if 04 gt 0 the amplitude of the oscillation is increasing as t a 00 and if 04 lt 0 the amplitude of oscillation is decreasing as t a 00 Also when 6 is large the functions have a smaller period of oscillation they oscillate more frequently Likewise if 6 is small the period of oscillation is greater the functions oscillate less frequently See Sections 34 and 35 for more practice 0 One should understand how to use the method of undetermined coef cients for non homogeneous linear di erential equations of any order I don7t foresee using any di erential equations of order larger than four on the exam Recall that given a di erential equation d y an71t dnily d W a1lt0 aoty W dt we start by solving the corresponding homogeneous di erential equa tion d d 1 d V L 71 y y y 7 at7 ati at 0 dtn n 1mm1 1dt oy Then we make an educated guess based on our gt as to a solution y t of the non homogeneous di erential equation If our edu cated guess contains functions that solve the homogeneous equation7 we multiply our educated guess by enough powers oft to avoid this problem For instance7 consider the di erential equation dgy dgy dy 7767 11776 i6t 11 5 dt3 alt dt 9 The corresponding homogeneous equation has general solution yo ct Clet 0262 0365 Our educated guess as to a solution of the non homogeneous equation might be something like y A1t A2 Aget Notice that 761 11 is a rst degree polynomial in the variable t7 so we would guess that a similar rst degree polynomial A1t A2 should be suf cient to achieve 761 11 when substituted into our di erential equation Also notice that the various derivatives of t are simply 6t again So we might guess that substituting Aget into our di erential equation might allow us to nd a coef cient A3 that solves the non homogeneous di erential equation However7 we then realize that t is a solution to the homogeneous di erential equation Thus we multiply that term of our guess by t So we set y A1t A2 Agtet 3 as our educated guess77 and substitute it into the di erential equation taking all the proper derivatives This then lets us determine the coef cients A1 A2 and A3 For more practice7 see Sections 36 and 43 of the course text or the relevant sample problems on the course website One should also know how to solve non homogeneous linear equations using variation of parameters One should recall that this method requires an understanding of the Wronskian from Sections 32 and 33 Because these require computing determinants7 the exam will not require one to solve any equation of order larger than 3 for such a problem For example7 one could be asked to nd the general solution to the di erential equation 2 g am d wmn Recall that we rst nd solutions to the corresponding homogeneous equation If ye c Oi 1 02 2 is such a solution then the solution to the non homogeneous equation is y 45W UiWWlW Z2 2 996 296 where m Ww mn Recall that W gt1 2x is the Wronskian of 1x and 2z and is given by the formula WWW det M ma 7 9 1 d Cl U2ltgt i d 02 45W 45 This material is covered in Section 37 An analogous result for higher order equations appears in Section 44 The method and formulas are similar except somewhat more complicated
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