MAT01B1: Separable Differential Equations

Transcription

1 MAT01B1: Separable Differential Equations Dr Craig 3 October 2018

2 My details: Consulting hours: Tomorrow 14h40 15h25 Friday 11h20 12h55 Office C-Ring (Or, just google Andrew Craig maths.)

3 Saturday classes There will be a Saturday class this week from 09h00 12h00. Venue will be announced on Blackboard.

4 Up until now we have often dealt with the situation where we know the following: dy dx = f(x). We can apply integration in order to obtain y = F (x) + C where F (x) is an antiderivative of f(x). There are many instances where we don t have an explicit formula for y, but we do have a formula for y (or y ) that uses y.

5 Population growth: Let P (t) be the population at time t. In some situations, the growth rate of a population is proportional to the size of the population. This gives rise to the (differential) equation dp dt = kp where k is some constant.

6 Population growth: If we solve we get dp dt = kp P (t) = Ce kt. This is called a general solution to a differential equation. When looking at applications then one is interested in getting a particular solution. This will often depend on the initial conditions.

7 The logistic model: in general, having P (t) proportional to P (t) ignores many important factors in population growth. As populations grow, resources become scarce and so it is often hard for a large population to grow as fast as a smaller population. The symbol M is often used to denote the carrying capacity of a population. As the population approaches M, the rate will decrease and if P (t) > M we should expect P (t) < 0.

8 The logistic model The following equation takes into account the carrying capacity M: ( dp dt = kp 1 P ) M If P is much smaller than M, we get P (t) kp. If P > M then P (t) < 0.

9 Hooke s Law: d 2 x dt 2 = k m x

10 Harmonic motion Consider a simple pendulum: The motion of a pendulum is measured by the equation of simple harmonic motion.

11 Harmonic motion continued... The equation of simple harmonic motion is a second-order differential equation: d 2 y dt 2 + ω2 y = 0. Here acceleration is a constant negative multiple of the displacement from a point of equilibrium. It is called a second-order DE because the it involves the second derivative of y(t). One way of expressing the solution is y(t) = A cos(ωt) + B sin(ωt)

12 Suppose that we solve a differential equation and obtain y = f(t) + C. If we think in terms of the graph of the function, the role of the constant C is to shift the graph vertically. When we get a specific solution (as opposed to a general solution) for a differential equation, what we are doing is to make sure that the graph of y = f(t) + C goes through the point (t 0, y 0 ).

13 Separable equations A separable equation is a first-order differential equation where dy can be dx written as a product of a function of x multiplied by a function of y. That is, dy = g(x) f(y) dx The challenge now is to solve this differential equation so that we get y as function of x.

14 Separable equations continued It will often be the case that we are given a differential equation in the form dy dx = g(x) h(y) This is still a separable equation, because we can rewrite it as dy dx = g(x) f(y) where f(y) = 1 h(y)

15 Solving separable equations From dy dx = g(x) h(y) we get h(y) dy = g(x) dx and then integrate: h(y) dy = g(x) dx From this point we can sometimes solve explicitly for y in terms of x.

16 Example: (a) Solve the differential equation dy dx = x2 y 2. (b) Find the solution that satisfies the initial condition y(0) = 2.

17 Example: Solve the differential equation Solution: dy dx = 6x 2 2y + cos y y 2 + sin y = 2x 3 + C

18 Example: Solve the equation y = x 2 y. We can rewrite this as dy y = x2 dx for y 0 (Remember that we have done this when we compute our final solution.)

19 Example: consider the differential equation L di + RI = E(t). dt L=inductance, I=current, R=resistance, E=voltage Find an expression for the current where the resistance R = 12Ω, inductance L = 4H, and a constant voltage of 60V. (Note: H = henries is the unit of measurement for inductance.)

20 Current example: Solving gives us 4 di dt + 12I = 60 I(t) = Ae 3t. Now find the limiting value of the current. lim I(t) = lim (5 t t 5e 3t ) = 5

21 Mixing problems: We can use differential equations to model the situation where a substance is being mixed in a tank. Here we will have dy = (rate in) (rate out) dt

22 Example: A tank contains 20kg of salt dissolved in 5000L of water. Brine containing 0.03kg/L of salt enters the tank at 25L/min. The solution is mixed thoroughly and drains from the tank at the same rate. How much salt remains in the tank after half an hour? Solution: y(t) = e t/200. y(30) = e 30/200

23 Graph of the solution:

24 More examples: 1. dp dt = P t, P (1) = 2 2. dy dθ = ey sin 2 θ y sec θ