1. Lesson 9-1 Differential Equations

2. Objectives Identify the order of a differential equation Test to see if a given solution is a solution to a differential equation Use Newton’s Law of Cooling to solve problems

3. Vocabulary Differential Equation – an equation that involves a derivative Order – the highest derivative involved in the differential equation Solution – a function that satisfies the differential equation Initial condition – allows the user to find the particular solution from a family of solutions Equilibrium – a steady state condition with neither growth nor decay

4. Differential Equations We call any equation that involves derivatives of any order a differential equation. Equations like the following are first-order: Equations like the following are second-order: Order then is simply the highest derivative involved in the equation. dy ---- = x + 1 dx dy = (x + 1) dx 0 = (x + 1) x’ – y’ d²y dy ---- = y + (x + 1) ------ dx² dx 0 = y + (x + 1) y’ – y’’

5. 9-1 Example 1 Is y = e -2t a solution to ? d²y dy ----- - 2 ----- - 8y = 0 dx² dx y = e -2t y’ = -2e -2t y’’ = 4e -2t Then “plug” into the equation and see if it satisfies the equation 4e -2t – 2(-2e -2t ) – 8(e -2t ) = 0  4e -2t - (-4e -2t ) - 8 e -2t = 0 8 e -2t - 8 e -2t = 0 so y = e -2t is a solution! First take the derivatives of the “solution”

6. 9-1 Example 2 dP P ----- = 0.03P ( 1 - ------- ) dt 2100 A population of wrens is governed by 1)For what values of P is the population increasing? 2)For what values of P is the population decreasing? 3)What are the equilibrium solutions? dP/dt > 0 if (1-p/2100) is positive, so if p < 2100 dP/dt 2100 dP/dt = 0 if (1-p/2100) is 0, so if p = 2100 or if p=0!

7. Solving Simple First-order Differential Equation Example: 1. Separate the Variables: get all x terms on one side and all y terms on the other. 2. Integrate both sides of the equation. 3. Combine the integration constants on the x side 4. Use any initial conditions to solve for C; like y(0) = 4 dy ---- = x dx dy = x dx ∫ dy = ∫ x dx  y + C = ½x² + C y = ½x² + C 4 = ½(0)² + C  4 = C y = ½x² + 4

8. 9-1 Example 3 The general solution to is y = Ce x², where C > 0. Find the solution if the initial value is y(1) = 3. dy ----- = 2xy dx y = Ce -x² y(1) = 3 = Ce -1² 3 = Ce -1 = C/e 3e = C = 8.1549

9. CSI: Newton’s Law of Cooling The rate at which an object cools is proportional to the difference in temperature between the object and the surrounding medium: Where T is temperature of the object, k is a proportionality constant and M is the temperature of the surrounding medium. A coroner uses this to help determine the time of death and is seen in every “Crime” TV series from Dragnet to CSI. dT ---- = k(T – M) dt

10. 9-1 Example 4 Example: A potato is taken out of a 300 o F oven and left to cool in a room at 75 o F. Write a differential equation expressing the change in rate of the temperature of the potato, T, with respect to time, t. dT ---- = k(T – M) dt dT ---- = k(300 – 75) dt dT ---- = 225k dt

11. 9-1 Example 4 cont Example: A potato is taken out of a 300 o F oven and left to cool in a room at 75 o F. Write a differential equation expressing the change in rate of the temperature of the potato, T, with respect to time, t. dT o ---- = k(T o – T m ) dt T(t) = T m + (T o – T m )e –kt T(t) = 75 + (300 – 75)e –kt T(t) = 75 + 225e –kt

12. Newton’s Law of Cooling Example: The great detective Sherlock Holmes and his assistant, Dr. Watson, are discussing the murder of actor Cornelius McHam. McHam was shot in the head, and his understudy, Barry Moore, was found standing over the body with the murder weapon in hand. Let’s listen in: Watson:Open-and-shut case, Holmes. Moore is the murderer. Holmes:Not so fast, Watson – you are forgetting Newton’s Law of Cooling! Watson:Huh? Holmes:Elementary, my dear Watson. Moore was found standing over McHam at 10:06 p.m., at which time the coroner recorded a body temperature of 77.9°F and noted that the room thermostat was set to 72°F. At 11:06 p.m. the coroner took another reading and recorded a body temperature of 75.6°F. Since McHam’s normal temperature was 98.6°F, and since Moore was on stage between 6:00 p.m. and 8:00 p.m., Moore is obviously innocent. Ask any calculus student to figure it out for you. How did Holmes know that Moore was innocent?

13. CSI Solution T(t) = T m + (T o – T m )e –kt T(t) is temperature of the body at t hours since death T m = 72 temperature of the room T 0 = 98.6 temperature of the body t would represent the hours since death But we don’t know the time of death. We can use the coroner's temperature readings to determine k. T(10:06) = 77.9 T(11:06) = 75.6 so T(1) = 75.6 = 72 + (77.9 – 72)e –k 3.6 = 5.9e –k k = -ln(3.6/5.9) =.494019 T(t=0) = 98.6 T(t, 10:06) = 77.9 = 72 + 26.6e -0.494019t 5.9 = 26.6e -0.494019t t = ln(5.9/26.6) / -0.494019 = 3.048 hours since death so death occurred at about 7 pm.

14. Summary & Homework Summary: –Differential equations involve derivatives –Order is determined by highest derivative involved in equation –Newton’s Law of Cooling has a wide variety of uses Homework: –pg 591 – 593: 5, 9, 10, 14