Section 3.1 Second Order Linear Homogeneous DEs with Constant Coefficients

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1 Section 3. Second Order Linear Homogeneous DEs with Constant Coefficients Key Terms/ Ideas: Initial Value Problems Homogeneous DEs with Constant Coefficients Characteristic equation

2 Linear DEs of second order are of crucial importance in the study of differential equations for two main reasons. The first is that linear equations have a rich theoretical structure that underlies a number of systematic methods of solution. Further, a substantial portion of this structure and of these methods is understandable at a fairly elementary mathematical level. Another reason to study second order linear equations is that they are vital to any serious investigation of the classical areas of mathematical physics. One cannot go very far in the development of fluid mechanics, heat conduction, wave motion, or electromagnetic phenomena without finding it necessary to solve second order linear differential equations.

3 An IVP involving a second order linear DE has the form This is just notation for a numerical value of the slope. Note that there are two initial conditions, one for y(t) and the other for y'(t). Observe that the initial conditions for a second order equation identify not only a particular point (t 0,y 0 ) through which the graph of the solution must pass, but also the slope y' 0 of the graph at the point (t 0,y 0 ). It is reasonable to expect that two initial conditions are needed for a second order equation because, roughly speaking, two integrations are required to find a solution and each integration introduces an arbitrary constant. Presumably, two initial conditions will suffice to determine values for these two constants.

4 SPECIAL TYPE OF SECOND ORDER LINEAR DEs In this chapter we will concentrate our attention on equations in which the coefficients are all constants. So the second order linear homogeneous equation with constant coefficients has the form ay" + by' + cy = 0 where a, b, and c are specified constants. It turns out that ay" + by' + cy = 0 can always be solved easily in terms of the elementary functions of calculus. On the other hand, it is usually much more difficult to solve a general second order DE if the coefficients are not constants. P(t)y" +Q(t)y' + R(t)y = 0

5 Before developing a general way to solve ay" + by' + cy = 0, let us first gain some experience by looking at a simple example that in many ways is typical. Example: Solve the IVP y" - y = 0, y(0) =, y'(0) = -. Observe that looking at the general form ay" + by' + cy = 0 we have a =, b = 0, and c = -. Rewrite the DE as y" = y. We want a function or functions whose second derivative is the same as the original function. From calculus two such functions that work are y (t) = e t and y (t) = e -t. But these are not the only ones. A little experimentation shows that any constant multiple of these work. For instance In addition, any sum or difference of solutions to this DE is another solution like WHY? Recall the superposition principle for linear homogeneous DEs.

6 Let us summarize what we have done so far in this example. For DE y'' y = 0 we have a pair of solutions y (t) = e t and y (t) = e -t and it follows that the linear combination y(t) = C e t + C e -t is also a solution. Since the coefficients C and C are arbitrary, this expression represents an infinite family of solutions (called the general solution) of the DE y'' y = 0. Now apply the initial conditions to determine C and C ; suppose that y(0) =, y'(0) = -. y(0) =, C + C = y'(t) = C e t - C e -t y'(0) = - C - C = - Thus we have the system of two equations C + C = C - C = - The solution is C = / and C = 3/. It follows that the solution of the IVP y" - y = 0, y(0) =, y'(0) = - is

7 General strategy for solving second order linear homogeneous DEs with constant coefficients ay" + by' + cy = 0 In our example we saw that exponential functions played and importanole. Step. We begin by assuming a solution of the form y = e rt. Step. Substitute y = e rt into the DE ay" + by' + cy = 0 a (e rt )'' + b(e rt )' + c(e rt ) = 0 a r (e rt ) + b r (e rt ) + c (e rt ) = 0 (e rt ) (a r + b r + c) = 0 Since e rt is never zero we have that ar + br + c = 0. Substitute. Differentiate. Factor. This is called the Characteristic Equation of the DE ay" + by' + cy = 0 Step 3. Find the roots of the characteristic equation ar + br + c = 0. Two roots r and r. Cases: both real and not equal both real but equal complex conjugates of one another

8 For each case there are techniques for getting the solution of the DE. Here we consider the case where the roots are r and r both real and not equal. Then we claim that the general solution of DE ay" + by' + cy = 0 in this case is Proof: Show that when we compute the Wronskian of it is not zero. r t and y (t) = e y (t) = e r t By definition the Wronskian is W = y (t) y' (t) - y' (t) y (t). We get r t d r t d r t r t dt dt r t r t r t W = e e - e e = e r e - r e e r t = r - r e e or equivalently r t e e r t e e W det det d r t d r t r t e e r e re dt dt = r t e r e - r e e r - r e e None of the factors can be zero so W 0 and hence the two solutions are said to be independent and we have a fundamental set of solutions. Thus the general solution is

9 Example: Solve IVP y'' + 5y' + 6y = 0, y(0) =, y'(0) = 3. Start by writing down the characteristic equation and finding its roots. Char. Eqn: r + 5r + 6 = 0 ; this factors as (r + )(r + 3) = 0, so the roots are r = - and r = -3. So the general solution of the DE is y(t) = C e -t + C e -3t. Now apply the initial conditions to determine C and C. y(0) = C + C = y'(0) = 3 -C e -*0-3C e -3*0 = -C -3C = 3 C + C = -C - 3C = 3 The solution is C = 9 and C = -7. It follows that the solution of the IVP y"+5y'+6y = 0, y(0) =, y'(0) = 3 is y(t) = 9e -t 7e -3t.

10 So IVP y"+5y'+6y = 0, y(0) =, y'(0) = 3 has solution y(t) = 9e -t 7e -3t. Describe the solution as t. y t t y(t) = 9e -t 7e -3t

11 Example: Solve IVP 4y''- 8y' + 3y = 0, y(0) =, y'(0) = /. Start by writing down the characteristic equation and finding its roots. 4r - 8r + 3 = 0 This factors as (r 3)(r ) = 0. so the roots are 3/ and / Therefore the general solution of the differential equation is y(t) = C e 3t/ + C e t/ Applying the initial conditions, we obtain the following two equations for C and C : y(0) = C + C = y'(0) = / 3/C e 0 + /C e 0 = 3/C + /C = / It follows that the solution of the IVP 4y'' - 8y'+3y = 0, y(0) =, y'(0) = / is y(t) = -/e 3t/ + 5/e t/ C + C = 3/C + /C = / The solution is C = -/ and C = 5/. Describe the solution of the IVP as t.

12 The IVP 4y''- 8y' + 3y = 0, y(0) =, y'(0) = / has a solution that goes through the point (0, ) with a slope of 3. The graph of the solution looks like y t t By inspecting the solution y(t) = -/e 3t/ + 5/e t/ we should have been able to anticipate this behavior.

13 The behavior of the solution of ay" + by' + cy = 0 is based on the roots of the characteristic equation. If the roots of the characteristic equation are real with r r, its general solution is the sum of two exponential functions. So the solution has a relatively simple geometrical behavior as t increases. The magnitude of the solution tends to zero when both roots of the characteristic equation are negative. If one root of the characteristic equation is positive the magnitude (absolute value) of the solution grows rapidly. This could be to or - depending upon the signs of C and C. There is also a third case that occurs less often: the solution approaches a constant when one root is zero and the other is negative.