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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

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