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Differential Equations for Applications

by: Mary Veum

Differential Equations for Applications MATH 3410

Marketplace > University of Connecticut > Mathematics (M) > MATH 3410 > Differential Equations for Applications
Mary Veum
GPA 3.97

Richard Bass

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Richard Bass
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This 7 page Class Notes was uploaded by Mary Veum on Thursday September 17, 2015. The Class Notes belongs to MATH 3410 at University of Connecticut taught by Richard Bass in Fall. Since its upload, it has received 29 views. For similar materials see /class/205803/math-3410-university-of-connecticut in Mathematics (M) at University of Connecticut.

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Date Created: 09/17/15
Math 3410 notes fall 2009 1 Some review 11 First order equations We will need to know how to do separable equations and linear equations A separable equation is one like dy 2 7 t dt y We can rewrite this as d i tdt 9 Integrating both sides7 1 it C y 2 7 so 1 y 17 it c A linear equation is one like 2 it 9 It i One multiplies by an integrating factor 2 pgft21ntt2 to get tzy Qty F or tsz t3 which leads to 2 7 1 4 t y 7 it c and then y it i 4 t239 When the linear equation has constant coef cients and is homogeneous ie7 the right hand side is 07 things are much easier To solve y 7 4y 0 we guess a solutions of the form y 6quot so y requot Then re 7 4e 07 or r 4 and therefore the solution is so we then have For non hornogeneous equations7 such as y 7 4y 63 one way to solve it is to solve the homogeneous equation y 7 4y 07 and then use 7 4t 9 i 05 yin where yp is a particular solution One way to nd a particular solution is to make an educated guess If we guess yp A65 then we have y 7 4 3A63t 7 4A 7 and this will equal 63 if A 71 We conclude the solution to the non hornogeneous equation is y 664 7 63 12 Series From calculus we have the Taylor series 2 3 5 1E 7 2 4 cosx17 gi and 7 3 5 Sln i 3X 7 lf 239 V71 substituting and doing some algebra shows that e cos z39 sinx 2 Second order linear 21 Applications First consider a spring hung from the ceiling with a weight hanging from it Let u be the distance the weight is below equilibrium There is a restoring force upwards of amount ku by Hooke7s law There is darnping resistance against the motion which is iRu And the net force is related to accelera tion by Newton7s laws so ku 7 RM F mu This leads to mu Ru 7 ku 0 If there is an external force acting on the spring then the right hand side is replaced by The second example is that of a circuit with a resistor inductance coil and capacitor hooked up in series Let I be the current Q the charge R the resistance L the inductance and C the capacitance We know that I dQdt The voltage drop across the resistor is IR across the capacitor QC397 and across the inductance coil L So if Et is the potential put into the current7 l Et LQ RQ 5Q Sometimes this is differentiated to give 1 E t LI RI 51 22 Linear constant coef cients homogeneous Let7s look at an example y 7 5y 4y 0 From Math 2117 we know a way of solving this Let 1 y and this one equation becomes a system 711 1 51 74y We then set up matrices7 where and the equation is X 7 AX We assume W w1 w2 and that our solution is of the form X We for some 7 wl and LU2 We will review this method later when we want to generalize it7 but lets look at an easier method 4 Let7s guess our solution is of the form for some 7 Then So our equation becomes e r2 7 57quot 4 0 This factors as r 7 4r 7 1 07 or r 14 The general solution to our equation is then y 0164 Cget This works except for two cases If 71 75 we do not get two solutions Thus7 for example7 the equation y 74y 4y0 leads to r 2 In this case we have as a solution y 0162 CgtEZt We di erentiate to check that this works Secondly we might have 71 a Mfg a 7 in Actually in this case we proceed and everything works out eventually It goes like this eltabilt catch e tcos bt z sin bt7 and similarly for 6014quot Then y C1614l7it C26a7ln39t 01th cos bt 0126 sin bt 026 cos bt 7 0226 sin bt 01 02km cos bt 01239 7 02239 6 sin bt dle cos bt dgeatabt Once we see how this goes7 we immediately can go to the last line without going through the derivation 23 Constants To get the values of 01 and 02 we need extra information In an initial value problern7 we are given yt0 and y t0 for some to For exarnple7 consider 2 4y 0 110 3710 4 We solve r2 4 07 or r i22 So y 01 cos 2t 02 sin 2t Then 3y00110207 or 01 1 Di cerentiating7 yt 7201 sin 2t 202 cos 2t7 and substituting in t 07 we get 02 2 One could also look at the same ODE but instead of initial values7 suppose we are given boundary values y0 0y1 3 Then as before 01 07 which leads to y 02 sin 2t Putting in t 17 we get 02 3sin 2 There is no analog of boundary value problems for rst order equations 24 Method of undetermined coef cients When solving y 7 5y 4y 6317 the general solution is h yp where yh is the general solution to the ho rnogeneous equation and yp is a particular solution to the non hornogeneous equation One way to get a particular solution is variation of parameters try y A65 Then y3A63t7 y9A63t7 and substituting gives 9A63t 7 15A63t 463 egt So A 7 and yp 763t and thus 4t t 1 3t ycle 626 756 To nd 0102 one waits until one has the most general solution before using initial or boundary values If the right hand side were 64 this doesn7t work7 but one could try yp Atet to get a particular solution If on the right hand side there was cos t7 one needs to try Acost Bsint One could also have 64 cost 262 on the right hand side7 for exarnple7 and one nds a particular solution for each piece7 and then adds 25 Euler equation An equation like xzy 4xy 7 4y 0 is called an Euler equation Try y as a trial solution Then y Tr717 y T0 71r7239 Substituting7 rr 71x1 2 4zrzT 1 7 4x7 07 or 7 7 7147quot7407 or r 741 Then the general solution is y clz 4 02x1 Again there are variations when 71 r2 or when r is complex If 71 75 the solution is 01x7 02x7 lnx lf 7 a i M the solution is y 01 cosb ln s 02x sinb ln


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