First order differential Equations

Transcription

1 Chapter 10 First order differential Equations 10.1 What is a Differential Equation? A differential equation is an equation involving an unknown function and its derivatives. A general differential equation can contain second derivatives and higher derivatives of the unknown function, but in this course we will only consider differential equations that contain first order derivatives. So the most general differential equation we will look at is of the form = f(x, y). (10.1) Here f(x, y) is any expression that involves both quantities x and y. For instance, if f(x, y) =x 2 + y 2 then the differential equation represented by (10.1) is = x2 + y 2. In this equation x is a variable, while y is a function of x. Differential equations appear in many science and engineering problems, as we will see in the section on applications. But first let s think of a differential equation as a purely mathematical question: which functions y = y(x) satisfy the equation (10.1)? It turns out that there is no general method that will always give us the answer to this question, but there are methods that work for certain special kinds of differential equations. To explain all this, this chapter is divided into the following parts: Some basic examples that give us a clue as to what the solution to a general differential equation will look like. Two special kinds of differential equation ( separable and linear ), and how to solve them. How to visualize the solution of a differential equation (using direction fields ) and how to compute the solution with a computer using Euler s method. Applications: a number of examples of how differential equations come up and what their solutions mean Two basic examples Equations where the RHS does not contain y Which functions y = y(x) satisfy =sinx? (10.1) This is a differential equation of the form (10.1) where the function f that describes the Right Hand Side is given by f(x, y) = sin x. In this example the function f does not depend on the unknown function y. Because of this the differential equation really asks which functions of x have sin x as derivative? In other words, which functions are the antiderivative of sin x? We know the answer, namely y = sin x = cos x + C 121

2 122 CHAPTER 10. FIRST ORDER DIFFERENTIAL EQUATIONS where C is an arbitrary constant. This is the solution to the differential equation (10.1). This example shows us that there is not just one solution, but that there are many solutions. The expression that describes all solutions to the differential equation (10.1) is called the general solution. It contains an unknown constant C that is allowed to have arbitrary values. To give meaning to the constant C we can observe that when x =0we have y(0) = cos 0 + C = 1+C. So the constant C is nothing but C = y(0)+1. For instance, the solution of (10.1) that also satisfies y(0)=4has C =4+1=5, and thus is given by y(x) = cos x +5. We have found that there are many solutions to the differential equation (10.1) (because of the undetermined constant C), but as soon as we prescribe the value of the solution for one value of x, such as x =0, then there is exactly one solution (because we can compute the constant C.) The exponential growth example Which functions equal their own derivative, i.e. which functions satisfy = y? Everyone knows at least one example, namely y = e x. But there are more solutions: the function y =0also is its own derivative. From the section on exponential growth in math 221 we know all solutions to = y. They are given by y(x) =Ce x, where C can be an arbitrary number. If we know the solution y(x) for some value of x, such as x =0, then we can find C by setting x =0: y(0) = C. Again we see that instead of there being one solution, the general solution contains an arbitrary constant C Summary The two examples that we have just seen show us that for certain differential equations there are many solutions, the formula for the general solution contains an undetermined constant C, the undetermined constant C becomes determined once we specify the value of the solution y at one particular value of x. It turns out that these features are found in almost all differential equations of the form (10.1). In the next two sections we will see methods for computing the general solution to two frequently occurring kinds of differential equation, the separable equations, and the linear equations First Order Separable Equations By definition a separable differential equation is a diffeq of the form y (x) =F (x)g(y(x)), or = F (x)g(y). (10.1) Thus the function f(x, y) on the right hand side in (10.1) has the special form f(x, y) =F (x) G(y). For example, the differential equation = sin(x)( 1+y 2) is separable, and one has F (x) =sinxand G(y) =1+y 2. On the other hand, the differential equation = x + y is not separable.

3 10.3. FIRST ORDER SEPARABLE EQUATIONS Solution method for separable equations To solve this equation divide by G(y(x)) to get 1 = F (x). (10.2) G(y(x)) Next find a function H(y) whose derivative with respect to y is H (y) = 1 ( ) solution: H(y) = G(y) G(y). (10.3) Then the chain rule implies that the left hand side in (10.2) can be written as 1 G(y(x)) = H (y(x)) = dh(y(x)). Thus (10.2) is equivalent with dh(y(x)) = F (x). In words: H(y(x)) is an antiderivative of F (x), which means we can find H(y(x)) by integrating F (x): H(y(x)) = F (x) + C. (10.4) Once we have found the integral of F (x) this gives us y(x) in implicit form: the equation (10.4) gives us y(x) as an implicit function of x. To get y(x) itself we must solve the equation (10.4) for y(x). A quick way of organizing the calculation goes like this: To solve = F (x)g(y) we first separate the variables, G(y) = F (x), and then integrate, G(y) = F (x). The result is an implicit equation for the solution y with one undetermined integration constant. Determining the constant The solution we get from the above procedure contains an arbitrary constant C. If the value of the solution is specified at some given x 0, i.e. if y(x 0 ) is known then we can express C in terms of y(x 0 ) by using (10.4) Example We solve Separate variables and integrate to get Finally solve for z and we find the general solution Example: the snag in action dz dt =(1+z2 )cost. dz 1+z 2 = cos t dt, arctan z =sint + C. z(t) = tan ( sin(t)+c ). If we apply the method to y (x) =y,weget y(x) =e x+c. No matter how we choose C we never get the function y(x) =0, even though y(x) =0satisfies the equation. This is because here G(y) =y, and G(y) vanishes for y =0.

4 124 CHAPTER 10. FIRST ORDER DIFFERENTIAL EQUATIONS 10.4 Problems For each of the following differential equations - find the general solution, - indicate which, if any, solutions were lost while separating variables, - find the solution that satisfies the indicated initial values. = xy, y(2) = 1. + x cos2 y =0, y(0) = π x =0, y(0) = A. 1+y 4. y 2 + x3 =0,y(0) = A y2 =0, y(0) = A. +1+y2 =0, y(0) = A. + x2 1 =0, y(0)=1. y 10.5 First Order Linear Equations Differential equations of the form equation are called first order linear. + a(x)y = k(x) (10.1) The Integrating Factor Linear equations can always be solved by multiplying both sides of the equation with a specially chosen function called the integrating factor. It is defined by A(x) = a(x), m(x) =e A(x). (10.2) Here m(x) is the integrating factor. It looks like we just pulled this definition of A(x) and m(x) out of a hat. The example in shows another way of finding the integrating factor, but for now let s go on with these two functions. Multiply the equation (10.1) by the integrating factor m(x) to get By the chain rule the integrating factor satisfies m(x) + a(x)m(x)y = m(x)k(x). dm(x) = d ea(x) = A (x) }{{} =a(x) e A(x) = a(x)m(x). }{{} =m(x) Therefore one has dm(x)y = m(x) + a(x)m(x)y { } = m(x) + a(x)y = m(x)k(x). Integrating and then dividing by the integrating factor gives the solution y = 1 ( m(x) ) m(x)k(x) + C. In this derivation we have to divide by m(x), but since m(x) = e A(x) and since exponentials never vanish we know that m(x) 0, so we can always divide by m(x).

5 10.5. FIRST ORDER LINEAR EQUATIONS An example Find the general solution to the differential equation Then find the solution that satisfies = y + x. y(2)=0. (10.3) Solution We first write the equation in the standard linear form Then we can see that and Thus, the integrating factor is y = x. a(x) = 1 k(x) =x. m(x) =e A(x), A(x) = a(x) = ( 1) = x + C. We only choose one integrating factor. The simplest choice is C =0. The solution to the differential equation is y(x) = 1 m(x)x m(x) = 1 e x e x x (integrate by parts) = e x{ } e x x e x + C = x 1+Ce x. This is the general solution. To find the solution that satisfies not just the differential equation, but also the initial condition (10.3), i.e. y(2) = 0, we compute y(2) for the general solution, y(2) = 2 1+Ce 2 = 3+Ce 2. The requirement y(2)=0then tells us that C =3e 2. The solution of the differential equation that satisfies the prescribed initial condition is therefore y(x) = x 1+3e x An example = y tan x +1, Solution Since We have Then Let m(x) =cosx,wehave y(x) = 1 m(x) a(x) = tan x, k(x) = 1. sin x tan x = = ln cos x + C cos x m(x) =e tan x =cosx + C. m(x)k(x) = 1 cos x With the initial condition y(0)=0, we conclude that C =0. cos x = sin x + C cos x.

6 126 CHAPTER 10. FIRST ORDER DIFFERENTIAL EQUATIONS 10.6 Problems 1. For each of the following differential equations specify the differential equation that the integrating factor satisfies, - find one integrating factor, - find the general solution, - find the solution that satisfies the specified initial conditions. In these problems K and N are constants. = y + x, =2y + x2, +2y + ex =0. (cos x)y = esin x,y(0) = A = 10y + e x, = y tan x +1, 8. cos 2 x = N y 9. x = y + x, = xy + x3, = y +sinx, = Ky +sinx, + x2 y =0, y(2)=0. y(1)=5. +(1+3x2 )y =0, y(1)=1.