DEFINITIONDIFFERENTIALEQUATIONS A differential equation is an equation dyinvolving a differential coefficient i.e.dx In this syllabus, we will only learn the firstFIRST ORDER DIFFERENTIAL EQUATIONSorder To solve differential equationdy, wedxintegrate and find the equation y whichsatisfies the differential equation1Exampledy x 1dxGeneral Solutiony y Particular Solution2Exampledy x 3x2dxx 1 dxx2 x c2General Solutiony Particular Solutionx 3x2 dxy x2 x 32y x23x3 c 23y x2 x 152y x2 x3 c2y x2 x 302y x2 x3 72y x2 x3 123Exampledy 1 xdxGeneral Solutiony y Particular Solution1 x dxSEPARATING THEVARIABLES1(1 x) 2 dx32y (1 x) c3 ( 1)23 2(1 x) 2y c34HOW DO WE KNOW IF IT CAN BE SEPARATED?3 2(1 x) 2y 2033y 2(1 x) 2 100356
To put it simply, you separate the variables of y and xxy2 dydxy on the leftdy 3dxdyx(y 1) 3dx 1(xy x)x on the rightdydx 1xy 2 dy 1dxx(y 1)dydx 3xy 2 dy 1dxx(y 1) dy 3dxx(y 1) dy y2 To put it simply, you separate the variables of y and x y on the left x on the right 3dxx78To put it simply, you separate the variables of y and xdy 5dxdyy(yx 1) 5dx(y 2 x y)y on the leftDIFFERENTIALEQUATIONSx on the rightCANNOT BE SEPARATEDSOLVING FIRST ORDER DIFFERENTIAL EQUATIONS910Example 1: Find the general solution of the differentialequationsMETHOD1.(a) Integrate the left hand side in terms of ySeparate the variables x and ydy1 dxx1dxx 11 dy dxx1 dy and the right hand side in terms of xdy g(x)dx f (y) dy g(x) dxf (y)y ln x c1112
Example 1: Find the general solution of the differentialequations1.(b)Example 1: Find the general solution of the differentialequations1.(c)dy cosec ydxdy 5ydx1dy 1 dxcosec y11sin y1dy 5 dxy 1dy 5 dxydy 1 dxsin y dy 1 dx sin y dy ln y 5x c1 dx cos y x c13Example 1: Find the general solution of the differentialequations1.(d)y(1 x)14Example 1: Find the general solution of the differentialequationsdy 2(1 y 2 ) 0dxy(1 x)1.(e)dy 2(1 y 2 )dxdy y2dx1 dy1 y 2 dxxydy2 (1 y 2 ) dx(1 x)11dy dxy2x 1y 2 dy dxxy2dy dx(1 y 2 )(1 x) y2dy dx(1 y 2 )(1 x)yxy 1 ln x c 1 ln(1 y 2 ) 2 ln(1 x) c2y1 ln x cy15Example 2: Solve the following differential equations:2.(a)dy 3x2 4x3dx 1 dy 3x2 4x3 dx 1 dy 3x2 4x3 dxy 3x34x4 c34y x3 x4 cExample 2: Solve the following differential equations:2.(b)y 1, x 1343416dy 5ydxy x x c 1 1 1 c1 2 c1dy 5 dxy 1dy 5 dxyy 2, x 0ln y 5x cln 2 5(0) cc ln 2ln y 5x cc 2 1ln y 5x ln 2c 1y e5x ln 2y x3 x4 1y e5x eln 2y 2e5x1718
Example 2: Solve the following differential equations:2.(c)xdy y2dxy 2 dy 1 ln x cy1 ln 1 c1 y3 3c 1 1 ln x cyy 1, x 0 y311 x sin 2x c324cos 2x 1 2 sin2 x1 1 cos 2x2 2sin2 x 1 1 cos 2x dx2 211 sin 2xy3 x c22 231 ln x 1yy dysin2 x dxy2y 2 dy sin2 x dx y 2 dy sin2 x dx 1 0 c1dxxy 1 ln x c 1 2.(d)x 1, y 11 dy1 dxy2x 11dy dxy2x Example 2: Solve the following differential equations:y311 x sin 2x c324 1ln x 11311 (0) sin 2(0) c3241 0 0 c31c 3y3111 x sin 2x 32431920Example 2: Solve the following differential equations:2.(e)(x 1) dy ydxy 1 ,x 111dy dxy(x 1)ln y ln(x 1) c ln 1 ln(1 1) c1dy y1dx(x 1)ln y ln(x 1) cDIFFERENTIALEQUATIONS0 ln 2 cc ln 2APPLICATION ON DIFFERENTIAL EQUATIONSln y ln(x 1) ln 221APPLICATION ON DIFFERENTIALEQUATIONS22APPLICATION ON DIFFERENTIALEQUATIONS Rate of change of a quantity Q isExamples are real lifeproportional to the value QdQαQdtNaturalGrowth Rate of change is taken with respect to timedQdt23NaturalDecayNewton’sLaw ofCoolingRate ofLiquidLeaking24
(1) Natural GrowthIf Q is increasing with time, it can be expressed asdifferential equationAPPLICATION ONDIFFERENTIALEQUATIONSdQαQdtdQ kQdtSolve the first order differential equation:NATURAL GROWTHANDNATURAL DECAYQ to the left . t to the right1dQ k dtQ 1dQ k dtQQ ekt cQ ekt ecQ Aektln Q kt cRate of increase of number of cells of yeast is proportional to the number of cells2526(II) Natural DecayMETHODIf the rate of decrease of Q is proportional to Q,dQ kQdtdQαQdt Interpret the word problem into aSolve the first order differential equation:Q to the left . t to the right1dQ k dtQ 1dQ k dtQdifferential equation Solve the first order of differential equation Find the value of k and A (or c) Answer the questionQ e kt cQ e kt cQ Aee ktln Q kt c27Example 2(1): The rate of increase of a colony of bacteria is proportional tothe number of bacteria present at any particular time. If initialnumber of bacteria is 100 and 4 hrs later there are 1350, thenfind k and A.28Example 2(1): The rate of increase of a colony of bacteria is proportional tothe number of bacteria present at any particular time. If initialnumber of bacteria is 100 and 4 hrs later there are 1350, thenfind k and A.B AektLet bacteria be B.dB α BdtdB kB k is positive becausedtit is a rate of increase 100 Aek01dB k dtB 1dB k dtB100 Ae0A 100k and A ?1350 100e4k1350 e4k100e4k 1.354k ln(1.35)k ln B kt cB e1350 Ae4kAt t 0, B 100At t 4, B 1350ln(1.35)4kt ck 0.0750B Aekt2930
Example 2(2): The rate of increase of a population of tigers is proportional tothe population at any particular instant of time. If initialnumber of tigers is 400 and if after 2 years there are 480, howmany tigers will there be after 5 years?Example 2(2): The rate of increase of a population of tigers is proportional tothe population at any particular instant of time. If initialnumber of tigers is 400 and if after 2 years there are 480, howmany tigers will there be after 5 years?Q AektLet tigers be Q.dQα QdtAt t 0, Q 400At t 2, Q 480dQ kQdtk is positive becauseit is a rate of increaseQ 400eln(1.2)t2400 Aek01dQ k dtQ 1dQ k dtQQ 400e400 Ae0480 Ae2kQ ekt ck 480 e2k400ktQ 631 tigers2k ln(1.2)480 400e2kln(1.2)(5)2Q 630.97A 400ln Q kt cQ AeAt t 5, Q ?ln(1.2)2e2k 1.231Example 2(3): The number of insects in a population t days after the start ofobservations is denoted by N. The variation in the number of insects ismodelled by a differential equation of the formdN kN cos(0.02t)dtExample 2(3): (ii)Given also that N 166 when t 30, find the value of k.N 125ekwhere k is a constant and N is taken to be a continuous variable. Itis given that N 125 where t 0.N Ae1dN k cos(0.02t) dtN 1dN k cos(0.02t) dtNln N k166 125ek125 Aesin(0.02t) c0.02N eksin(0.02t)0.02sin(0.02(0))0.02ln 1.328 k125 Ae0A 125k N 125eksin(0.6)0.02sin(0.6)166 ek 0.021251.328 ekksin(0.02(30))0.02166 125eksin(0.02t)0.02At t 0, N 125sin(0.02t) c0.02N ekksin(0.02t)0.02At t 30, N 166(i) Solve the differential equation, obtaining a relation between N, k and t.dN kN 02calculator inradians!(ln 1.328)(0.02)sin 0.6k 0.0100479eck 0.01003334Example 2(3): (iii)Obtain an expression for N in terms of t, and find the least value of Npredicted by this model.N 125e0.0100sin(0.02t)0.02k 0.0100479APPLICATION ONDIFFERENTIALEQUATIONSk 0.0100 1 sin t 1 1 sin(0.02t) 1 1(0.0100) 0.0100 sin(0.02t) 1(0.0100) 1(0.0100)sin(0.02t)1(0.0100) 0.0100 0.020.020.02 0.502 0.0100sin(0.02t) 0.5020.02N 125eNEWTON’S LAW OF COOLING 0.502N 75.66N 763536
Example 3(1): A cup of coffee, originally at 90 C is left to cool in a room at aconstant 20 C . After 5 minutes, the temperature is 80 C . Findhow long the coffee takes to cool to 60 C .NEWTON’S LAW OF COOLINGThe rate of cooling of a body is proportional to the difference between it’s temperature and thetemperature of the surroundingsAt t 0Initial Body temperatureLet θ be the temperature of the bodyLet θ0 be the temperature of the surroundingsdθdtα (θ θ0 )ln(θ θ0 ) kt ln 7090 CAt t 5, θ 80 CSurrounding temperature 20 Cdθ k(θ θ0 )dtdθdtdθ k(θ θ0 )dtα (θ θ0 )ln(80 20) k(5) ln 70ln 60 5k ln 701dθ k dtθ θ0 1dθ k dtθ θ01dθ k dtθ θ0 1dθ k dtθ θ05k ln 60 ln 70k k 0.0308ln(θ θ0 ) kt cln(θ θ0 ) kt cln 60 ln 705ln(90 20) k(0) cln(θ θ0 ) 0.0308t ln 70c ln 7037Example 3(1): A cup of coffee, originally at 90 C is left to cool in a room at aconstant 20 C . After 5 minutes, the temperature is 80 C . Findhow long the coffee takes to cool to 60 C .38Example 3(2): A liquid is heated in an oven kept at a constant 180 C . It is assumed thatthe rate of increase in the temperature of the liquid is proportional to(180 C θ) where θ is the temperature of the liquid at time t minutes.If the temperature rises from 0 C to 120 C in 5 minutes, find thetemperature of the liquid after a further 5 minutes.ln(θ θ0 ) 0.0308t ln 70 ln(180 0) k(0) cAt t 0Initial Body temperature 0 C At t ?, θ 60 Cc ln(180)Surrounding temperature 180 C ln(180 θ) kt ln 180 Surrounding temperature 20 Cdθdtln(60 20) 0.0308t ln 70 α (180 C θ)At t 5, θ 120dθ k(180 θ)dtln 40 0.0308t ln 700.0308t ln 70 ln 40 ln(180 120) k(5) ln 180 ln 60 5k ln 1801dθ k dt(180 θ) 1dθ k dt(180 θ)ln 70 ln 40t 0.0308t 18.25k ln 180 ln 60ln 180 ln 605k 0.220k ln(180 θ) kt c ln(180 θ) 0.220t ln 18039Example 3(2): A liquid is heated in an oven kept at a constant 180 C . It is assumed thatthe rate of increase in the temperature of the liquid is proportional to(180 C θ) where θ is the temperature of the liquid at time t minutes.If the temperature rises from 0 C to 120 C in 5 minutes, find thetemperature of the liquid after a further 5 minutes.Further 5 minutest 10θ ?40Example 3(3): The temperature of a quantity of liquid at time t is θ . The liquid iscooling in an atmosphere whose temperature is constant and equal to A. Therate of decrease of θ is proportional to the temperature difference (θ A).Thus θ and t satisfy the differential equation.dθ k(θ A)dtwhere k is a constant.(i) Find, in any form, the solution to this differential equation, given that θ 4A when t 0 ln(180 θ) 0.220t ln 180dθ k(θ A)dt ln(180 θ) 0.220(10) ln 180ln(180 θ) 0.220(10) ln 180 180 θ e 0.220(10) ln 180θ 180 e 0.220(10) ln 1801dθ k dt(θ A) 1dθ k dt(θ A)ln(θ A) kt cθ 180 20ln(4A A) k(0) cc ln 3Aθ 160ln(θ A) kt ln 3A4142
Example 3(3): (ii)Given also that θ 3A when t 1, show that k ln32Example 3(3): (iii) Find θ in terms of A when t 2, expressing your answer in its simplest form.ln(θ A) kt ln 3Aln(θ A) At t 1, θ 3AAt t 2ln(3A A) k(1) ln 3Aln(θ A) ln ln3232 t ln 3A(2) ln 3Aθ A e( ln 2 )(2) ln 3A3ln 2A k ln 3Ak ln 3A ln 2Aθ e( ln 2 )(2) ln 3A A33Ak ln2Aθ e( ln 2 )(2) eln 3A A33k ln23θ e( 2 ln 2 ) (3A) Aθ e ln( 32 ) 2 (3A) A4θ e(ln 9 ) (3A) Aθ 4(3A) A9θ 4(A) A3θ 7A343APPLICATION ONDIFFERENTIALEQUATIONS44ON THE BOARD :DRATE OF LIQUID LEAKING4546
DIFFERENTIAL EQUATIONS FIRST ORDER DIFFERENTIAL EQUATIONS 1 DEFINITION A differential equation is an equation involving a differential coefﬁcient i.e. In this syllabus, we will only learn the ﬁrst order To solve differential equation , we integrate and ﬁnd the equation y which
(iii) introductory differential equations. Familiarity with the following topics is especially desirable: From basic differential equations: separable differential equations and separa-tion of variables; and solving linear, constant-coefﬁcient differential equations using characteristic equations.
Andhra Pradesh State Council of Higher Education w.e.f. 2015-16 (Revised in April, 2016) B.A./B.Sc. FIRST YEAR MATHEMATICS SYLLABUS SEMESTER –I, PAPER - 1 DIFFERENTIAL EQUATIONS 60 Hrs UNIT – I (12 Hours), Differential Equations of first order and first degree : Linear Differential Equations; Differential Equations Reducible to Linear Form; Exact Differential Equations; Integrating Factors .
3 Ordinary Differential Equations K. Webb MAE 4020/5020 Differential equations can be categorized as either ordinary or partialdifferential equations Ordinarydifferential equations (ODE's) - functions of a single independent variable Partial differential equations (PDE's) - functions of two or more independent variables
Chapter 1 Introduction 1 1.1 ApplicationsLeading to Differential Equations 1.2 First Order Equations 5 1.3 Direction Fields for First Order Equations 16 Chapter 2 First Order Equations 30 2.1 Linear First Order Equations 30 2.2 Separable Equations 45 2.3 Existence and Uniqueness of Solutionsof Nonlinear Equations 55
Introduction to Advanced Numerical Differential Equation Solving in Mathematica Overview The Mathematica function NDSolve is a general numerical differential equation solver. It can handle a wide range of ordinary differential equations (ODEs) as well as some partial differential equations (PDEs). In a system of ordinary differential equations there can be any number of
13.1 Differential Equations and Laplace Transforms 189 13.2 Discontinuous Functions 192 13.3 Differential Equations with Discontinuous Forcing 194 Problem Set E: Series Solutions and Laplace Transforms 197 14 Higher Order Equations and Systems of First Order Equations 211 14.1 Higher Order Linear Equations 212
DIFFERENTIAL – DIFFERENTIAL OIL DF–3 DF DIFFERENTIAL OIL ON-VEHICLE INSPECTION 1. CHECK DIFFERENTIAL OIL (a) Stop the vehicle on a level surface. (b) Using a 10 mm socket hexagon wrench, remove the rear differential filler plug and gasket. (c) Check that the oil level is between 0 to 5 mm (0 to 0.20 in.) from the bottom lip of the .
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