By Peter V. O'Neil

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Here is an example that suggests why this might be useful. 20 The equation y2 − 6xy + 3xy − 6x2 y = 0 is not exact on any rectangle. 12) have the same solution. 12) is just y y2 − 6xy + 3xy − 6x2 y = 0 and if y = 0, then necessarily y2 − 6xy + 3xy − 6x2 y = 0. 11) as well. To review what has just occurred, we began with a nonexact differential equation. We multiplied it by a function chosen so that the new equation was exact. We solved this exact equation, then found that this solution also worked for the original, nonexact equation.

6) The function e p x dx is called an integrating factor for the differential equation, because multiplication of the differential equation by this factor results in an equation that can be integrated to obtain the general solution. 6). Instead, recognize the form of the linear equation and understand the technique of solving it by multiplying by e p x dx . 14 The equation y + y = sin x is linear. Here p x = 1 and q x = sin x , both continuous for all x. An integrating factor is e dx or ex . 15 Solve the initial value problem y = 3x2 − y x y 1 =5 First recognize that the differential equation can be written in linear form: 1 y + y = 3x2 x An integrating factor is e to get 1/x dx = eln x = x, for x > 0.

Then T − 68 = ±Aekt = Bekt Then T t = 68 + Bekt Now the constants k and B must be determined, and this requires information. m. and immediately measured the body temperature, obtaining 94 4 degrees. Letting 9:40 be time zero for convenience, this means that T 0 = 94 4 = 68 + B and so B = 26 4. Thus far, T t = 68 + 26 4ekt To determine k, the lieutenant makes another measurement. At 11:00 she finds that the body temperature is 89 2 degrees. Since 11:00 is 80 minutes past 9:40, this means that T 80 = 89 2 = 68 + 26 4e80k Then e80k = 21 2 26 4 so 80k = ln 21 2 26 4 16 CHAPTER 1 First-Order Differential Equations and k= 21 2 1 ln 80 26 4 The lieutenant now has the temperature function: T t = 68 + 26 4eln 21 2/26 4 t/80 In order to find when last time when the body was 98 6 (presumably the time of death), solve for the time in T t = 98 6 = 68 + 26 4eln 21 2/26 4 t/80 To do this, the lieutenant writes 30 6 = eln 21 2/26 4 t/80 26 4 and takes the logarithm of both sides to obtain ln 30 6 26 4 = 21 2 t ln 80 26 4 Therefore the time of death, according to this mathematical model, was 80 ln 30 6/26 4 ln 21 2/26 4 t= which is approximately −53 8 minutes.