Differential Equation For Dummies j91p

Differential Equations for Dummies Colin Mayo AP Calculus Croatan High School

Table of Contents Slide 3. What is a differential equation? Slide 4. Example – Newton’s Law of Cooling Slide 5. Differential Equation – A model Slide 6. Graphical solution – Slope Field

Slide 7. Analytical solution – Separation of Variables Slide 9. Numerical solution – Euler’s Method Slide 10. Works cited

What is a Differential Equation? • Any equation that includes a rate of change • Usually contains a derivative Examples: dH/dt = -k(H -20) dP/dt = 2P-2T dy/dx = -y/x

Example: Newton’s Law When a murder is committed, the body, originally at 37°C, cools according to Newton’s Law of Cooling. Suppose that after two hours the temperature is 35°C, and that the temperature of the surrounding air is a constant 20°C. 1. Find the temperature, H, of the body as a function of t, the time in hours since the murder was committed. 2. If the body was found at 4 pm at a temperature of 30°C, when was the murder committed?

Differential Equations: A Model dH/dt = -.063(H - 20) The change in body temperature with respect to time will equal a constant times the difference between the body and the surrounding air temperature

Solution: Slope Field • Solution can be found graphically • Particular solution can be found using initial value • Using initial value of (37,0) and trace shows t is around 8.5 when H is 30. • Therefore time of death was approximately 8.5 hours before 4 pm or around 7:30 am.

Solution: Separation of Variables • Solution can found analytically • Separating variables gives equation for family of solutions • Particular solution can be found using initial value

dH/dt = -.036(H-20)

dH = -.036(H-20)dt 1/(H-20) dH = -.036dt ∫1/(H-20) dH = ∫ -.036dt ln(H-20) = -.036 t + C H-20 = e-.036 t + C H-20 = A e-.036 t H(t) = A e-.036 t + 20

H(0) = 37 = A e-.036 (0) + 20 37 = A + 20 A= 17 H(t) = 17 e-.036 t + 20

Solution: Separation of Variables • General Solution can be used to find when the body temperature is 30 degrees. • Time of death was approximately 8.4 hours prior to 4 pm which is around 7:30 am.

H(t) = 17 e-.036 t + 20 30 = 17 e-.036 t + 20 30 – 20 = 17 e-.036 t 10 = 17 e-.036 t 10/17 = e-.036 t ln(10/17) = -.036 t

ln(10/17)/-.036 = t t = 8.4 hours

Solution: Euler’s Method • Solution can be found numerically • Euler’s Method gives a particular solution • ynew = ynew + ∆x(y’) • Since 30 occurs between 8 and 8.5, time of death occurred at approximately 8 hours before 4 pm which would be 7:30 am

H

Hnew

0

37

.5

36.4645

1

35.94586825

1.5

35.4435734

2

34.95710084

2.5

34.4859216

3

34.02964467

3.5

33.58771086

4

33.15969797

4.5

32.74516748

5

32.34369471

5.5

31.95486832

6

31.57828997

6.5

31.21357384

7

30.86034626

7.5

30.51824535

8

30.18692063

8.5

29.86603263

Works Cited Hughes-Hallett, Deborah, et al. Calculus. 3rd ed. Hoboken: John Wiley & Sons, 2002.