8.a: Integrating Factors in Differential Equations. y = 5y + t (2)

Transcription

1 8.a: Integrating Factors in Differential Equations Basics of Integrating Factors Until now we have dealt with separable differential equations. Net we will focus on a more specific type of differential equation, that is first order, linear ordinary differential equations or first order linear ODEs for short. What does this very long name mean? First order means that only y and y appear in our differential equation. There are no second, third, or fourth derivatives, (y, y, y (4) ) etc. For eample y = 5y + t (1) is first order but, is not. y = 5y + t (2) ordinary differential equations are differential equations that are only in terms of a single variable. If you have more than one variable then you are dealing with partial differential equations which are much harder and which we will not work with in this course. The most important term here for our purposes is linear. A differential equation is linear when it can be written in the form y + p()y = q(). (3) Eample 0.1. Solve the differential equation y = 6y + 6. (4) We already know one method for solving this via separation of variables. We rewrite the equation as y = 6(y + 1) (5) and separate variables to get Net we can integrate to get Solving for y gives y = 6. (6) y + 1 ln y + 1 = C. (7) y = C e 32 1 (8) Let s try a different method though. Let s multiply both sides of the equation by e 32 to get e 32 y = 6e 32 y + 6e 32 (9) Moving the 6e 32 y to the left side then gives, e 32 y 6e 32 y = 6e 32 (10) Notice that the left side is the derivatives of e 32 y using the product rule. Thus when we integrate both sides we get e 32 y 6e 32 y = 6e 32 = e 32 y = e 32 + C. (11) Solving for y gives, Notice that the two answers agree. y = 1 + Ce 32 (12) 1

2 This second method that we used is called the integrating factor method. In this case the integrating factor was the e 32 that we multiplied the equation by. For a general first order, linear ODE, we go through the following steps to solve using integrating factors y + p()y = q() (13) 1. Compute the antiderivative p()d. Call this P (). We define our integrating factor to be 2. Multiply the entire equation by µ so that we get µ = e P (). (14) e P () y + p()e P () y = e P () q(). (15) 3. Notice that e P () y + p()e P () y = (e P () y) (16) 4. Integrate both sides to get (e P () y) = e P () q() = e P () y = e P () q() (17) 5. Solve for y by dividing both sides by e P (). To know when we can use integrating factors we need to make sure that we can identify linear first order ODEs. Eample 0.2. Decide if each is linear and if so give p() and q(). 1. y y = 0 2. y y = 3. y 2 y 2 = 0 4. y y 2 = 2 Solutions: 1. Linear, p() =, q() = Linear, p() = 1, q() =. 3. Not linear, there is a factor of y Not linear, there is a factor of y 2. Now that we have a procedure for solving linear, first order ODE s lets look at some more eamples Eample 0.3. Solve the differential equation To solve this we use our list of steps from above: y + y =. (18) 1. In this equation p() = 1 so that P () =. Then µ = e. 2. We multiply the entire equation by µ = e to get e y + e y = e. (19) 3. As usual, e y + e y = (e y) (20) 2

3 4. Integrating both sides gives (e y) = e = e y = e e + C. (21) 5. Dividing both sides by e gives, Net lets try a harder eample Eample 0.4. Solve the differential equation To begin with we need to get this into the form (13). So we divide by y = 1 + Ce. (22) y + ( + 1)y = e sin(2), > 0. (23) y ( + 1)y + = e sin(2). (24) This makes the right side look like it is going to be nasty to integrate but for now lets continue and see what happens. We go through our usual steps. 1. Notice that so integrating this we get Thus 2. Multiplying both sides by e gives 3. We write p() = + 1 = 1 + 1, (25) P () = + ln(). (26) µ = e P () = e +ln() = e (27) e y + ( + 1)e y = sin(2). (28) e y + ( + 1)e y = (e y). (29) 4. Integrating both sides of our equation gives (e y) = sin(2) = e y = cos(2) + C. (30) 2 5. Solving for y gives, y = e cos(2) 2 Eample 0.5. Solve the equation y cos() + y sin() = 1, π 2 < < π 2. + Ce. (31) Again we need to put this in the form (13). To do this we divide the equation by cos() to get Net we go through our steps: 1. In this eample p() = tan(). Integrating this gives P () = ln sec() so 2. We multiply both sides by sec() to get 3. We write 4. So integrating both sides gives (sec()y) = y + tan()y = sec(). (32) µ = e P () = e ln sec() = sec(). (33) sec()y + tan() sec()y = sec 2 (). (34) sec()y + tan() sec()y = (sec()y). (35) sec 2 () = sec()y = tan() + C. (36) 5. Dividing both sides by sec() give y = sin() + C cos() (37) 3

4 0.1 Single Compartment Model When rate of water entering and leaving is the same We now look at a basic application for linear, first order ODE s: the single compartment model. The simplest version of this consists of one compartment, for eample a fied volume of water V containing a solute. Assume that water enters and leaves the compartment at the same rate so that the total volume stays the same. We are interested in how the concentration of solute will change over time. Of course this will depend on what the concentration of solute entering the compartment is. Our notation will be the following V is the volume of water. C(t) is the concentration of solution in the compartment at time t. It follows that the total mass of solute in the compartment is V C(t). C I is the concentration of the incoming solution. q is the rate at which water enters the compartment (and therefore also leaving since we are assuming these are the same). So qc I is the rate at which solute enters the compartment. If the solution is well mied then the rate at which solute is leaving the compartment will be qc(t). The rate of change of solute mass will then be: Rate of change of mass in solute in compartment = (rate at which mass enters) (rate at which mass leaves). (38) So our differential equation is then d dt (CV ) = qc I qc. (39) Since V is constant then we can rewrite this as dt = q V (C I C). (40) This a differential equation that we can solve! Especially when we change the form slightly to get Now lets look at an eample. dt + q V C = q V C I. (41) Eample 0.6. Suppose that a solution containing 7 grams per gallon of pollutant flows into a 1000-gallon tank at a rate of 2 gallons per minute, and the well-stirred miture flows out of the tank at a rate of 2 gallons per minute. The tank was initially full with no pollution. Write the differential equation for C(t) and solve it. Solution: Firstly, lets identify the relevant terms. V = 1000 gal q = 2 gal/min C I = 7 g/gal Then plugging these into our equation we get dt + qc V Using the integrating factor e.002t we get = qc I V = dt +.002C =.014 (42).002t e dt +.002e.002t C =.014e.002t (43) 4

5 Integrating both sides (since we already have a C variable, I am going to use B) should give e.002t C = 7e.002t + B = C(t) = 7 + Be.002t. (44) Of course we need an initial condition if we are to find B. In this case the initial condition was the statement that the tank initially had no pollution. So that C(0) = 0. Plugging this in, we find B = 7 so that our final solution is C(t) = 7 7e.002t (45) Also, we would epect that given an infinite amount of time the concentration in the tank would be equal to the incoming concentration. Indeed, we see lim t C(t) = 7, so this agrees with our epectations. 5