Introduction to Differential Equations

Transcription

1 > 22M:16 Fall 05 J. Simon ##################### Introduction to Differential Equations NOTE: This handout is a supplement to your text. There are several homework problems, part of the HW that is due on Thursday Sept 29. The problems are labeled DiffEq #1, DiffEq #2, etc. ##################### Suppose y=f(x), where f(x) = x^2. You know how to calculate the derivative, dy/dx = f ' (x) = 2x. What if the question were reversed: Find a function y=f(x) for which dy/dx = 2x. This is an example of a differential equation, that is some equation involving a function and/or some of its derivatives. If someone says, "Solve the differential equation...", they are asking you to FIND a function y=f(x) that satisfies the given equation. Often when people set up mathematical model of "real world" situations, what they set up are differential equations. Example 1. "The acceleration of gravity is constant" is the statement that if some object is falling under the effect of gravity, and we use y=f(t) to denote the position at time t, then the second derivative f ''(t) is constant. The equation is > Diff(Diff(y,t),t)= - 32 ; > 2 t 2 y = -32 To "solve" this equation, we need to find a function y=f(x) that has the property that f ''(x) = -32. For example, the function y= 16 x^ x - 10 is a solution. Here is how you would check that: > y:=16*x^2 + 17*x - 10; y := 16 x x - 10 > dydx:=diff(y,x); dydx := 32 x + 17 > d2ydx2:=diff(dydx,x);

2 d2ydx2 := 32 You calculate the second derivative of y with respect to x, and observe that it really is 32. Example 2. Check that the function y = x^3 satisfies the differential equation dy/dx = 3 y/x. Solution: If y=x^3 then dy/dx = 3 x^2. On the other hand, 3 y/x = 3 x^3 / x = 3 x^2, which is what we found for dy/dx. In pretty Maple printing, this would be... > y:=x^3; y := x 3 > dydx:=diff(y,x); > ThreeYoverX:=3*y/x; > is(dydx=threeyoverx); dydx := 3 x 2 ThreeYoverX := 3 x 2 true ############################# HOMEWORK DiffEq #1 Verify that the function y=5x^3 satisfies the differential equation dydx = 3y/x. ############################# As you see from the above homework problem, a given differential equation may have more than one "solution". In fact, typically, a differential equation has infinitely many solutions. Example 3. Solve the differential equation dy/dx = 10. We seek a function y=f(x) such that dy/dx = 10. So we are looking for a function whose graph has constant slope. Any line of slope 10 will satisfy this equation. SO f(x) = 10x or f(x) = 10x + 3 or f(x) = 10x - 17 or f(x) = 10x - sqrt(73) are all solutions to the given differential equation. In fact, any function of the form f(x) = 10 x + C, where C is a constant, is a solution. It turns out there are no other solutions (need more calculus to prove that). We call y = 10x + 34 a "particular solution" of the differential equation dy/dx = 10, and y = 10x + C the "general solution".

3 Example 4. Find the general solution of the differential equation dy/dx = x. [NOTE: We have not yet developed the machinery to know when we have **all** possible solutions. So for now, the phrase "general solution" will just mean an infinite family of solutions corresponding to one or more constants that can be chosen arbitrarily.] Solution for Example 4: We want a function y=f(x) for which dy/dx = x. Since d(x^2)/dx = 2x, we see that y=(1/2) x^2 has the right property. But we also know that the derivative of a constant function is 0, so if we add any constant to (1/2) x^2, we won't change the derivative; it will still be "x". SO... One particular solution is y=(1/2) x^2 The general solution is y = (1/2) x^2 + C, where C is an arbitrary constant. ###################### HOMEWORK DiffEq2 Find one particular solution of the differential equation dy/dx = 3x. Find the general solution. ###################### HOMEWORK DiffEq 3 Find one particular solution of the differential equation dy/dx = 2x + 3. Find the general solution. ######################## HOMEWORK DiffEq 4 Find one particular solution of the differential equation dy/dx = x^2 + 5x. Find the general solution. ########################### Describing a falling object. Suppose we throw a ball straight upwards, with an initial speed of 9 0 ft/sec. When will it hit the ground? How high will it get?

4 This is a very hard problem if you try to include the friction due to air-resistance. (Think of the complicated trajectories that baseballs, tennis balls, golfballs, footballs take when thrown; the interaction with the air is significant. See the web page for an interesting discussion.) To simplify the problem (so we'll just get an approximate answer), we will assume the ball is small and smooth and that it moves without air-resistance. Let y(t) denote the height at time t. Newton (F = ma) tells us that the second derivative d2y/dt2 = -32 (feet/sec)/sec. Let's view the derivative dy/dt = v as a function itself and first find what that is. Since d(dy/dt) / dt = -32, we know that dy/dt might be -32 t. It might also be -32 t , or any function of the form -32t + C. How do we decide?? Look at what happens at time t = 0. We are told the initial upward velocity is 90 ft/sec. So at time t = 0, dy/dt = -32(0) + C = 90. Thus C = 90. The velocity dy/dt must be -32 t Now we want to solve the following differential equation: Find y=f(t) such that dy/dt = -32 t This is just like previous examples and homework problems. We know that y = -16 t^ t would work. So would y = -16 t^ t work. The general solution is y = -16 t^ t + K, where K is any constant. So how do we find K? Again, look at what happens at time t=0. At time t=0, the height y is K. In fact, the person who stated this problem did not give enough information. To know the height at all times ( i.e. to know the function y(t) ), we need to know the height at t=0, the initial height. Let's say that the ball was launched with an initial height of 5 feet. So y(0) = K = 5. We now have the particular solution to the original differential equation y ' ' = -32 that satisfies not only the differential equation, but also the given "initial conditions" saying y(0) = 5 and y ' (0 ) = 90. The solution is y= -16 t^ t + 5. Now we can answer all sorts of questions. Let's first graph the height function. > y:=-16*t^2+90*t+5; Look at what happens at time t=0. y := -16 t t + 5 > plot(y, t=-2..10, color=blue, thickness=3);

5 Having plotted this function for t=-2..10, we now realize that some of this time period is physically meaningless. We launched the ball at t=0, so the behavior of the function for t<0 is not part of our physical experiment. Similarly, once the graph reaches t=6, we have y<0; but the ball has hit the ground, so the motion is no longer described by our simple equation. Nevertheless, seeing a lot of the parabola is useful because it reminds us that our function y=f(t) is a familiar quadratic, where we know how to find the zeros and we know how to find where the maximum is attained. > solve(y=0,t); , > evalf(%); , Remember, the ball was launched from height 5 ft, that is y(0) = 5. At time t = approx 5.7 seconds, the ball hits the ground. To find how high it got, recall that the vertex of the parabola occurs midway between the roots. We see from the above solution that is at t=45/16. How high is the ball at that moment? > subs(t=45/16,y); > evalf(%); So the ball reaches a height of a little over feet.

6 ################# HOMEWORK DiffEq 5 A ball is thrown vertically from an initial height of 4 feet, with an initial upwards velocity of 60 ft/sec. When does the ball hit the ground? How high does it get?? [You may ignore air resistance.] #################### Simple Differential Equations Force Us to Study Exponential and Trigonometric Functions In the next chapter, we will study exponential functions. You studied functions such as cos(x), ln(x), 2^x, etc. in high school. One of the main reasons for people to study these kinds of functions is that "Nature" forces us to. The physical world talks to us in the language of differential equations, and we need exponential and trig functions to solve even the simplest differential equations. We saw one important example above, that of a body moving freely under the force of gravity. There, the solutions were just [quadratic] polynomials. But consider the following situations. ############### Exponential growth ############### The simplest model of unrestricted population growth is to say that the rate of growth is proportional to the current population. This is saying that at any given moment, some fixed percent of the population is reproducing. If P(t) is the population at time t, we state our assumptions about the growth rate as a differential equation, for example dp/dt =.02 P The function ^ t is the (approximate, I'm rounding off an irrational number) solution if we add the initial condition that P(0)=1. > P:=1.0202^t; > diff(p,t); P := t t ##################### Simple spring #################### Suppose you have some object at the end of a spring; you stretch out the spring and let go. Describe the motion of the object.

7 For a typical spring, the longer you stretch it, the harder it pulls back. The simplest model of this is to say that the force pulling back is proportional to the amount you stretch the spring. Springs also resist compression, and the simplest model is, again, that the force the spring exerts is proportional to how much you compress it away from its relaxed position. Let x(t) denote the displacement at time t (so x=0 is the neutral, rest position). Newton (again, F=mA), together with our assumption that force is proportional to displacement, say that the acceleration is proportional to the displacement, but in the opposite direction: x ' ' (t) = - k x(t), where k is some positive constant. Sines and cosines give us the solutions. > X:=A*cos(sqrt(k)*t) + B*sin(sqrt(k)*t); X := A cos( k t) + B sin( k t) > Vel:=diff(X,t); > Acc:=diff(Vel,t); Vel := -A sin( k t) k + B cos( k t) k Acc := -A cos( k t) k - B sin( k t) k Notice that X ' ' is just (-k) times X.