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Survey of Calculus II

by: Donny Graham

Survey of Calculus II MTH 126

Donny Graham
GPA 3.57

Andrew Cooper

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Andrew Cooper
Class Notes
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This 6 page Class Notes was uploaded by Donny Graham on Saturday September 19, 2015. The Class Notes belongs to MTH 126 at Michigan State University taught by Andrew Cooper in Fall. Since its upload, it has received 73 views. For similar materials see /class/207296/mth-126-michigan-state-university in Mathematics (M) at Michigan State University.

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Date Created: 09/19/15
MTH 126 NOTES ON SEPARATION OF VARIABLES ANDREW AI COOPER 1 INTRODUCTION So far we have seen two kinds of differential equations which we know how to solve We know that equations of the form dy 7 t dt f gt are solved by yt Ft7 where F is the antiderivative of f We7ve also veri ed that dyi i 7 k dt 9 is solved by yt OE and that dy 7 k i A dt 1 is sovled by yt A 06 In addition we saw that the solution of dy y 7 k 1 7 7 dt y L is yt Other than simply guessing7 how do we come up with solutions to equations For example7 we might want to nd a solution of cry yz dt The technique for doing this is called separation of variables It relies on the chain rule 2 THE CHAIN RULE AND INTEGRATION BY SUBSTITUTION Recall that the Chain Rule dz dz dy dx Eyds allows us to treat differential symbols like jig as fractions 1 2 ANDREW A COOPER The Chain Rule when integrated gives Integration by Substitution mowwelt Fltglttgtgt o where F is the antiderivative of f 3 SEPARATION OF VARIABLES Suppose that our differential equation has a right hand side that looks like a function of t times a function of y For example we might have dy 7 dt 7 If yt is a solution of this equation and yt 31 0 we can divide both sides by yt2 to get tyz dig it t W Now both sides are functions of It so we can integrate them with respect to t 12 dt idt tdt W Let7s rewrite 12 as yt 2 and as y t yltt yt 2y tdt tdt Now the left hand side is ready for Integration by Substitution with gt yt and fg 9 2 So we integrate the left hand side ill t 01 tdt Now the right hand side is easy to integrate t2 yt 1 C1 E 02 We7ve succeeded in removing the differential symbols and we7re left with an equation involving y and t We want to write yt so lets solve algebraically First combine the constants together by C MTH 126 NOTES ON SEPARATION OF VARIABLES 02 701 1 mweg v owo 1 1 g0em mm m o 1 1 Rewrite this as mw4m Now lets check that this function actually solves the differential equation we7re interested in First compute dy d t2 7 1 aa m t2 3m4t Now compute the right hand side which we can see is the same as 2 So in fact the function yt 7g C 1 is a solution to the differential equation 7 tyz 4 SEPARATION OF VARIABLES USING FRACTIONS Remember we saw before that the Chain Rule Integration by Sub stitution they are really the same thing allows us to treat dy E as a fraction It turns out we can encode the reasoning we just used by doing exactly this 4 ANDREW A COOPER Start by multiplying both sides by dt dy dti t 2dt dt 9 1 2 1 Edy ty dtyj d 1 tdt 9 Now integrate both sides dy tdt 1 1 t2 7 0 70 1 1 2 2 and then continue with the same algebra as before Let7s try some other examples using fractions this way 1 For examples with the right hand side depending only on t this approach is simple and straightforward For exarnple dt 7 dy tzdt dy t2dt t3 901 02 t3 7 C 9 3 t2 MTH 126 NOTES ON SEPARATION OF VARIABLES 2 We also know how to solve jig ky dy ik dt 9 d kdt y d i kdt y 1ny 01 kt 02 lny kt 03 y 6kt03 y 6036 y CE 3 And to solve My 7 A dy k 7A dt 1 dy kdt yiA y kdt yiA dyA kdt yA lny7AClktCg lny7AktCg yiA6ktCg yACekt 6 ANDREW A COOPER 4 dyi 2 gig dy idt yz dy P 1 71 fly 01t02 71 it03 y 7 I yiOit 5 dy 3 E79 dy Edt dy E 1 397201t02 1 2 77 t O 29 3 y 2072t y072t 5 INITIAL CONDITIONS Recall that an Initial value problem lVP consists of two parts a differential equation7 and an Initial value So far all the formulas we7ve come up with have involved a con stant7 which we have called 0 This constant comes from the 0 in integrating both sides How do we determine it Suppose I want to solve the lVP y ya y0 10


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