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Section 6.2

Section 6.2. Differential Equations. A differential equation is an equation involving a derivative (slope). A solution to a differential equation is a function whose derivative is the differential equation. Example 1. Find the general solution to each differential equation. a . y ꞌ = 2.

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Section 6.2

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  1. Section 6.2 Differential Equations

  2. A differential equation is an equation involving a derivative (slope). • A solution to a differential equation is a function whose derivative is the differential equation.

  3. Example 1 • Find the general solution to each differential equation. • a. yꞌ = 2

  4. b. yꞌ + 2x2 = 0

  5. c. yyꞌ = -x

  6. A family of solutions is infinite. • A particular solution is unique.

  7. Example 2 • Determine whether the function y = x2ex – 5x2 is a solution of the differential equation • xyꞌ − 2y = x3ex. • Take the derivative of y = x2ex – 5x2.

  8. yes xyꞌ − 2y = x3ex

  9. Example 3 • Use integration techniques to find a general solution to

  10. Example 4 • Find the particular solution to the differential equation yꞌ = cosx – sin x given y() = -1.

  11. General Solution (, -1) Particular solution

  12. 2nd Day Differential Equations: Growth and Decay 6.2

  13. Growth and Decay Models In many applications, the rate of change of a variable y is proportional to the value of y. If y is a function of time t, the proportion can be written as follows.

  14. Solve the differential equation: So, all solutions of yꞌ = ky are in the form of y = Cekt

  15. Growth and Decay Models • If y is a differential function of t such that y > 0 • and yꞌ = ky, for some constant k, then • y= Cekt • C is the initial value of y, and k is the • proportionality constant. Exponential growth • occurs when k > 0, and exponential decay • occurs when k < 0.

  16. Example 1 • The rate of change of y is proportional to y. When t = 0,y= 2, and when t = 2, y = 4. What is the value of y when t = 3?

  17. Because y' = ky,you know that y and t are related by the equation y =Cekt. • You can find the value of the constant C by applying the initial conditions y = 2 when t = 0. • 2 = Ce0 • C = 2

  18. You can find the value of the constant k by applying the initial conditions y = 4 when t = 2. • 4 = 2e2k • 2 = e2k

  19. So, the model is y ≈2e0.3466t. • When t = 3, • y = 2e0.3466(3) • ≈ 5.657 • Notice that we did not actually have to solve the differential equation y' = ky, we just used its solution y =Cekt.

  20. The next example concerns Newton's Law of Cooling, which states that: • the rate of change in the temperature of • an object is proportional to the difference between the object’s temperature and the temperature of the surrounding medium. • y′ = k(TO− TM)

  21. Example 2 • Let y represent the temperature (in ºF) of an object in a room whose temperature is kept at a constant 60º. If the object cools from 100º to 90º in 10 minutes, how much longer will it take for its temperature to decrease to 80º?

  22. From Newton's Law of Cooling, you know that the rate of change in y is proportional to the difference between y and 60. • This can be written as • y' = k(y – 60), 80 ≤ y ≤ 100.

  23. Using y = 100 when t = 0, then • Because y = 90 when t = 10, • 90 = 60 + 40ek(10) • 30 = 40e10k

  24. So, the Cooling Model is: y = 60 + 40e–0.02877t • and finally, when y = 80, you obtain • So, it will require about 14.09 more minutes for the object to cool to a temperature of 80º

  25. Half-Life Formula • Radioactive material decays exponentially and is measured in terms of half-life (the number of years required for half of the atoms in a sample of radioactive material to decay).

  26. Example 3 • A sample contains 1 gram of radium. How much radium will remain after 1000 years? (Use a half-life of 1620 years.) • C = 1 grams at time t = 0.

  27. Example 4 • Suppose that 10 grams of the plutonium isotope Pu-239 was released in the Chernobyl nuclear accident. How long will it take for the 10 grams to decay to 1 gram? Pu-239 has a half life of 24,100 years. • y = Cekt

  28. Example 5 • Suppose a population of fruit flies increases according to the law of exponential growth (y = Cekt). There are 100 flies after Day 2 and 300 flies after Day 4. Approximately how many flies were in the original population?

  29. Solve the system by dividing.

  30. Example 6 • Money is deposited in an account for which • interest is compounded continuously. If the balance doubles in 6 years, what is the annual percentage rate? • A = Pert

  31. A = Pert • 2P = Pert • 2 = e6r • ln 2 = 6r

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