6.5 Separable Differential Equations and Exponential Growth

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1 Separable Differential Equations and Exponential Growth The Law of Exponential Change It is well known that when modeling certain quantities, the quantity increases or decreases at a rate proportional to the size of the quantity. If we let y 0 denote the amount present at time t = 0, then we obtain the following differential equation. (3) dt = ky, with y(0) = y 0 Here k is the growth (or decay) constant. Equation (3) is an example of a separable differential equation. More specifically, it is a separable initial value problem. Now if y 0 then (3) implies = ˆ 1 y 1 y dt = k ˆ dt dt = kdt = ln y = kt+c y = e kt+c y = e C e kt y = ±e C e kt = Ae kt Notice that y(0) = Ae k0 = A and we allow A = 0 whenever y = 0 then the solution to the initial value problem (3) is y = y(t) = Ae kt. We have the so-called Law of Exponential Change. (4) y = y 0 e kt where the rate constant k is called the growth constant if k > 0 or the decay constant if k < 0.

2 6.5 3 Remark. Be aware that some texts rewrite (4) when discussing exponential decay as In this case, k is always positive. y = y 0 e kt Population Growth Under certain practical assumptions population growth obeys the law of exponential change. (What are some of these assumptions?) Example 2. Bacteria Growth Suppose that a petri dish initially contains 3000 bacteria and that 12 minutes later there are 3500 bacteria. a. Find a formula for the bacteria population t hours (not minutes) after the initial measurement. We ll use the model y = y 0 e kt with y 0 = Now we must find the growth constant k. We know that y(12/60) = 3500 so we must solve the following equation for k. y(1/5) = 3500 = 3000e k1/5 = = ek1/5 ln = k 5 It follows that k = 5ln and our growth model is 30 (5) y(t) = 3000e t b. Predict the bacteria population in 4 hours. We use equation (5). y(4) = 3000e (4) 65472

3 6.5 4 Remark. Notice that 4 hours is twenty 12 minute intervals, so that y(4) = 3000 ( ) Continuously Compounded Interest Suppose that you invest P 0 dollars at a fixed annual interest rate r (expressed as a decimal). If interest is added (compounded) to your account n times per year, the formula for the value of that account after t years is (6) ( P(t) = P 0 1+ r ) nt n This is the so-called compound interest formula. Now suppose that the number of compounding periods (per year) is increased say from n = 12 (monthly), to n = 52 (weekly) or hourly, etc. In fact, what happens to our initial investment if we increase n without bound. Mathematically, we have ( lim P(t) = lim P 0 1+ r ) nt n n n = P 0 lim n (1+ r n ) nt Now let m = n/r. Since r is fixed, we observe that n m Thus ( lim P(t) = P 0 lim 1+ 1 ) mrt n m m [ ( = P 0 lim 1+ m) 1 m ] rt, (why?) m = P 0 e rt

4 6.5 5 This lead to Continuously Compounded Interest Formula (7) P(t) = P 0 e rt In this case the (growth) constant r is called the continuous (annual) interest rate. Newton s Law of Cooling Suppose that a hot potato is left to cool on a kitchen table. If we let H s equal the room temperature (the surrounding temperature) then one obtains the following differential equation. (8) dh dt = k (H H s) The solution to (8) leads to the following equation (9) H H s = (H 0 H s ) e kt called Newton s Law of Cooling. Example 3. Cooling a Hot Potato Immediately after a hot potato is removed from an oven its temperature is measured to be 350 F. Ten minutes later the potato s temperature is 300 F. If the room temperature remains constant at 68 F, how long will it take the potato to cool to 100 F? So H s = 68 and H 0 = 350. So by (9), hence = (350 68)e 10k = e 10k = k = ln(232/282)

5 6.5 6 We now have a model for the temperature of the potato t minutes after it was removed it from the oven. Now we set H(t) = 100 and solve for t. Thus H(t) = 68+(350 68)e t 100 = e t = e t = t ln(32/282) So it takes almost two hours for the potato to cool to 100 F?? General First-Order Equations Definition. (10) A first-order differential equation is an equation of the form dx = f(x,y) We say that y = y(x) is a solution to (10) if y(x) is defined on an interval I and d y(x) = f (x,y(x)) dx Example 4. a. Let A 0 be a real number. One can easily verify that y = Ae x3 is a solution to the differential equation (11) dx = 3x2 y.

6 6.5 7 b. Consider the differential equation (12) y = y 2 Observe that (13) (14) are solutions to (12). y = 1/(x+4) y = 1/(x+C) Equation (13) is called a particular solution and (14) is called a general solution. Remark. The adjective first-order refers to the number of derivatives under consideration. In later courses (e.g., Differential Equations), higher-order differential equations are discussed in more detail. For example, d 2 y dx 2 = xy3 is called a second-order differential equation. Separable Equations A differential equation is called separable if it can be written as (15) In such cases (15) can rearranged so that (16) dx = g(x)h(y) h(y) = g(x)dx Now we integrate both sides to obtainˆ (17) ˆ h(y) = g(x) dx which leads to a (implicit) solution to (15).

7 6.5 8 Example 5. Solve the differential equation below. (18) dx = y cos 2 y y cos 2 y = dx Now we integrate both sides. ˆ =ˆ y cos 2 y ˆ sec 2 ˆ y = y dx dx So the implict solution of (18) is given by 2tan y = x+c Can we solve for y explicitly in terms of x?