19.4. Integrating Factor. Introduction. Prerequisites. Learning Outcomes. Learning Style

Transcription

1 Integrating Factor 19.4 Introduction An exact differential equation is one which can be solved by simply integrating both sides. Whilst such equations are few and far between an important class of differential equations can be converted into exact equations by multiplying through by a function known as the integrating factor for the equation. In this Block you will learn how to decide whether an equation is capable of being transformed into an exact equation,how to determine the integrating factor and how to obtain the solution of the original equation. Prerequisites Before starting this Block you should... Learning Outcomes After completing this Block you should be able to... explain the purpose of an integrating factor determine the integrating factor for a first-order differential equation where appropriate solve a variety of equations using this technique 1 understand what is meant by a differential equation; (Block 19.1) 2 be familiar with the terminology associated with differential equations: order, dependent variable and independent variable; (Block 19.1) 3 be able to integrate; (Block 14) Learning Style To achieve what is expected of you... allocate sufficient study time briefly revise the prerequisite material attempt every guided exercise and most of the other exercises

2 1. How does an integrating factor work? The equation x 2 dy +3xy= x3 is not exact. However,if we multiply it by x we obtain the equation x 3 dy +3x2 y = x 4. This can be re-written as d (x3 y)=x 4 which is an exact equation with solution x 3 y = x 4 = 1 5 x5 + C and hence y = 1 5 x2 + C x 3. The function by which we multiplied the given differential equation in order to make it exact is called an integrating factor. In this example the integrating factor is simply x itself. Now do this exercise Which of the following differential equations can be made exact by multiplying by x 2? (a) dy + 2 x y = 4 dy (b) x +3y = x2 (c) 1 dy x 1 x y = x 2 (d) 1 dy x + 1 x y =3. 2 Where possible,write the exact equation in the form d (f(x) y) =g(x). Answer 2. Finding the integrating factor for linear differential equations The differential equation governing the current i in a circuit with inductance L and resistance R in series subject to a constant applied electromotive force E cos ωt where E and ω are constants is L di + Ri = E cos ωt (1) This is an example of a linear differential equation in which i is the dependent variable and t is the independent variable. The general standard form of a linear first-order differential equation is normally written with y as the dependent variable and with x as the independent variable and arranged so that the coefficient of dy is 1. That is,it takes the form: dy + f(x) y = g(x) in which f(x) and g(x) are functions of x. In our example x is replaced by t and y by i to produce di + f(t) i = g(t). The function f(t) is the coefficient of the dependent variable in the differential equation. We shall describe the method of finding the integrating factor for (1) and then generalise it to a linear differential equation written in standard form. Engineering Mathematics: Open Learning Unit Level 1 2

3 Step 1 Write the differential equation in standard form i.e. with the coefficient of the derivative equal to 1. Here we need to divide through by L: di + R L i = E cos ωt. L Step 2 Integrate the coefficient of the dependent variable (that is, f(t) R/L) with respect to the independent variable (that is, t),and ignoring the constant of integration R L = R L t. Step 3 Take the exponential of the function obtained in step 2. This is the integrating factor (I.F.) I.F. = e Rt/L. This leads to the following Key Point on integrating factors: Key Point The linear differential equation (written in standard form): [ dy + f(x)y = g(x) has an integrating factor I.F. = exp ] f(x) Now do this exercise Find the integrating factor for the equations (a) x dy +2xy= xe 2x (b) t di +2ti= te 2t (c) dy (tan x)y =1. Answer 3. Solving equations via the Integrating Factor Having found the integrating factor for a linear equation we now proceed to the solution of the equation. Returning to the differential equation,written in standard form: for which the integrating factor is di + R L i = E cos ωt L e Rt/L we multiply the equation by the integrating factor to obtain Rt/L di e + R L ert/l i = E L ert/l cos ωt At this stage you will find the left-hand side of this equation can always be simplified: d (ert/l i)= E L ert/l cos ωt. 3 Engineering Mathematics: Open Learning Unit Level 1

4 Now this is in the form of an exact differential equation and so we can simply integrate both sides to obtain the solution: e Rt/L i = E e Rt/L cos ωt. L All that remains is to complete the integral on the right-hand side. Using the method of integration by parts we find e Rt/L L cos ωt = [ωt sin ωt + R cos L 2 ω 2 ωt]ert/l + R2 Hence Finally e Rt/L i = E L 2 ω 2 + R 2 [ωt sin ωt + R cos ωt] ert/l + C. E i = L 2 ω 2 + R [ωl sin ωt + R cos ωt]+c 2 e Rt/L. is the solution to the original differential equation (1). Note that,as we should expect for the solution to a first-order differential equation,it contains a single arbitrary constant C. Now do this exercise Using the integrating factors found earlier find the general solutions to the differential equations (a) x 2 dy +2x2 y = x 2 e 2x (b) t 2 dy +2t2 y = t 2 e 2t (c) dy (tan x)y =1. Answer More exercises for you to try 1. Solve the equation x 2 dy + xy=1. 2. Find the solution of the equation x dy y = x subject to the condition y(1)=2. 3. Find the general solution of the equation dy + (tan t) y = cos t. 4. Solve the equation dy + (cot t) y = sin t. 5. The temperature θ (measured in degrees) of a body immersed in an atmosphere of varying temperature is given by dθ +0.1θ =5 2.5t. Find the temperature at time t if θ =60 when t =0. 6. In an LR circuit with applied voltage E = 10(1 e 0.1t ) the current i is given by L di + Ri = 10(1 e 0.1t ). If the initial current is i 0 find i subsequently. Answer Engineering Mathematics: Open Learning Unit Level 1 4

5 4. Computer Exercise or Activity For this exercise it will be necessary for you to access the computer package DERIVE. To solve an exact differential equation using DERIVE it is necessary to load what is called a Utility File named ode1. To do this is simple. Proceed as follows: In DERIVE,choose File:Load:Math and select the file (double click) on the ode1 icon. This will load a number of commands which enable you to solve first-order differential equations. You can use the Help facility to learn more about these if you wish. Of particular relevance here is the DERIVE command Integrating factor(p,q,x,y,x0,y0) which obtains the solution of a differential equation of the form q(x, y) dy + p(x, y) =0 y(x0) = y0 if the differential equation is exact when multiplied by an integrating factor. Use the command Integrating Factor Gen(p,q,x,y,c) to obtain a solution without initial conditions but which includes a single arbitrary constant c. For example,to solve x dy y = x y(1)=2 proceed as follows:- Choose Author:Expression,Integrating Factor( y x, x, x, y, 1, 2) and select Simplify. DERIVE responds with [x LN x +2 x = y] As a useful exercise use DERIVE to check the solutions given in the exercises presented above. Also note that all the differential equations presented in this Block are linear differential equations having the general form dy + p(x)y = q(x) y(x0) = y0 Such equations can also be solved in DERIVE using the command Linear1(p,q,x,y,x0,y0) or by using the related command Linear1 Gen(p,q,x,y,c) for a solution to a differential equation without initial conditions but which contains a single arbitrary constant c. Also use the Help command to find out about the more general commands Dsolve1(p, q, x, y, x0,y0) and Dsolve1 Gen(p, q, x, y, c) used for solving very general first-order ordinary differential equations. 5 Engineering Mathematics: Open Learning Unit Level 1

6 End of Block 19.4 Engineering Mathematics: Open Learning Unit Level 1 6

7 (a) Yes. x 2 dy +2xy=4x2 becomes d (x2 y)=4x 2. (b) Yes. x 3 dy +3x2 y = x 4 becomes d (x3 y)=x 4. (c) This equation is already exact as it can be written in the form d (d) Yes. x dy + y =3x2 becomes d (xy) =3x2. Back to the theory ( ) 1 x y = x. 7 Engineering Mathematics: Open Learning Unit Level 1

8 (a) Step 1 Step 2 Divide by x to obtain dy +2y =e 2x The coefficient of the independent variable is 2 hence 2 =2x Step 3 I.F. =e 2x (b) The only difference from (a) is that i replaces y and t replaces x. Hence I.F. =e 2t. (c) Step 1 This is already in the standard form. sin x Step 2 tan x = = ln cos x. cos x Step 3 I.F. =e ln cos x = cos x Back to the theory Engineering Mathematics: Open Learning Unit Level 1 8

9 (a) The standard form is dy +2y =e 2x for which the integrating factor is e 2x. Then,on multiplying through by e 2x we have 2x dy e +2e2x y = 1 i.e. d (e2x y) = 1 so that e 2x y = x + C leading to y = (x + C)e 2x (b) The general solution is y =(t + C)e 2t (c) The equation is in standard form and the integrating factor is cos x. d then (cos xy) = cos x so that cos xy= cos x = sin x + C giving y = tan x + C sec x Back to the theory 9 Engineering Mathematics: Open Learning Unit Level 1

10 1. y = 1 x ln x + C x. 2. y = x ln x +2x 3. y =(t + C) cos t 4. y = ( 1 t 1 sin 2t + C) cosec t 5. θ = t 240e 0.1t i = 10 ( ) [ ] 100 R e 0.1t 10L + i 0 + e Rt/L. 10R L R(10R L) Back to the theory Engineering Mathematics: Open Learning Unit Level 1 10