First Order Linear Ordinary Differential Equations
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1 First Order Linear Ordinary Differential Equations The most general first order linear ODE is an equation of the form p t dy dt q t y t f t. 1 Herepqarecalledcoefficients f is referred to as the forcing term in the equation. When f 0, we say the equation is homogeneous when f is not identically zero, we say the equation is inhomogeneous. Before trying to make any general statements about the solution of this equation, we will consider several examples. Examples of Homogeneous Equations 1. Consider p t 1, q t t, f t 0. Then y t ty t 0, dy y tdt. Then ln y t 1 2 t2 C 0 y t C 1 e t2 /2. This 1-parameter family of solutions exists is smooth for all values of t. For C 1 1, the graph looks like Figure 1 2. Consider p t 1, f t 0, q t to say y t C 0 if t 0. For t 0, y t ty t 0, 0 if t 0 t if t 0 y t C 1 e t2 /2. Then. Then for t 0, y t 0, which is y t C 0 if t 0 C 1 e t2 /2 if t 0. 1
2 This solution satisfies the equation for all t but the solution is continuous only when we choose C 0 C 1. With this choice we get the following graph Figure 2 3. Consider p t 1, f t 0, q t continuous solution is, 0 if t 0 t 1 if t 0. In this case we find the only y t C 1 if t 0 C 1 e t t2 /2 if t 0 The graph of this solution is, Figure 3 4. Consider p t t 1, q t 1, f t 0. In this case t 1 y t ty t 0, we get dy y 1 t 1 dt, which leads to ln y t ln t 1 C 0. The explicit solution is then or ln y t ln t 1 ln y t t 1 C 0 2
3 y t C 1 t 1. This solution is valid for t 1 or for t 1, but it fails to exist at t 1 hence it is not valid over any interval that contains t 1. Note that this example is equivalent to the example where p t 1, q t 1/ t 1, f t 0. We can make the following observations about these examples. All the equations were homogeneous so the f plays no role in any of these examples. In example 1, p q are both smooth functions the solution as seen in Figure 1 is smooth for all values of t. In example 2, the coefficient q is continuous but has a discontinuity in its derivative at t 0. This produces the solution seen in Figure 2, which is continuous with a continuous derivative for all t but at t 0, there is a discontinuity in the second derivative of the solution. In example 3, the coefficient q is discontinuous at t 0 this is seen to produce a solution, shown in Figure 3, which is continuous for all t but which has a discontinuity in the derivative of the solution at t 0. Finally, in example 4, the coefficient q t 1/ t 1, is undefined (we say q has a singularity) at the point t 1 this produces a solution that has a singularity at the same value of t. Evidently, a point where the coefficient p t 0 is equivalent to a point where the coefficient q t has a singularity. Accordingly, in the future, we will agree to rewrite the equation (1) in the form y t q t p t y t f t p t or y t a t y t F t, 2 admit the possibility that the coefficient a the forcing term F may have singularities. We can summarize the observations about these examples as follows. In considering the homogeneous version of equation (2) (i.e., F 0) we have seen that the smoothness of the solution is determined by the smoothness of the coefficient a t. When a t is smooth, the solution of the equation is smooth, but as a t becomes less smooth, the solution also loses some smoothness until, finally, when there is a point where the coefficient has a singularity, the solution may fail to exist. Solutions of Inhomogeneous Equations In all the examples above, the equation was homogeneous. Now consider equation (2) in the case where F is not zero so the equation is said to be inhomogeneous. In order to find a solution in this case we begin by first solving the corresponding homogeneous equation, y t a t y t 0. This leads to or dy y t a dt C 0 A t C 0 y H t C 1 e A t where A t a t. Here y H t is referred to as the general solution for the homogeneous equation. This means that for any choice of the constant C 1, the function y H t solves the homogeneous ODE, every solution of the homogeneous equation must be given by this formula for 3
4 some choice of C 1. The first half of this statement has been proved by the construction we have just performed. The second half of the statement is true but has not been proved yet. At any rate, we will now use the homogeneous solution to find a solution for the inhomogeneous equation. This is accomplished by supposing the inhomogeneous equation has a solution of the form y p t C t e A t. Werefertothisasaparticular solution. Here C t denotes an unknown function which we will now find. Note that y p t C t e A t C t e A t A t. C t e A t C t e A t a t. Then y p t a t y p t C t e A t C t e A t a t a t C t e A t C t e A t, the inhomogeneous equation now becomes, or y p t a t y p t C t e A t F t C t F t e A t. We integrate to get C t, C t t F e A d, then y p t C t e A t e A t t F e A d. This method of finding a particular solution is called the method of variation of parameters. The assumption that the particular solution could be written in the form y p t C t y H t had the effect of reducing the inhomogeneous equation to a simple integration to find C t. The general solution for the inhomogeneous equation is defined to be the sum of the general homogeneous solution a particular solution, y t y H t y p t C 1 e A t e A t t F e A d. Note that if we add any homogeneous solution to a particular solution, we obtain a new particular solution. It is for this reason that we refer to y p t as A particular solution not THE particular solution. We will illustrate this method with an example. Consider 4
5 y t ky t t k constant We find the homogeneous solution to be y H t Ce kt. Now we suppose the particular solution has the following form y p t C t e kt where Then C t e kt F t t C t F t e kt te kt, C t te kt dt y p t C t e kt 1 k 2 ekt kt 1, 1 2 kt 1 k Then the general solution for this example is y t Ce kt 1 2 kt 1. k Another way to obtain the particular solution is by guessing (or we could call it the method of undetermined coefficients). Here we note that since y p t ky p t t, it is reasonable to assume that into the equation, we find y p t at b where a,b are to be determined. Substituting at b k at b a kb akt t. Then, equating coefficients of like powers of t on the two sides of this last equation, or ak 1 a kb 0, a 1 k b a k 1 k 2. This leads to y p t at b 1 k t 1, which agrees with the previous result. 2 k As a second example, consider the inhomogeneous equation, 5
6 y t ky t cos t where k,, all denote given constants. If we use the method of variation of parameters we suppose y p t C t e kt where C t e kt F t cos t C t F t e kt e kt cos t. Then C t e kt cos tdt : e kt k cos t sin t, k 2 2 y p t C t e kt k cos t sin t. k 2 2 This particular solution could as easily have been obtained by guessing. Since the forcing term involves cos t, it is logical to suppose that the particular solution could only be composed of some combination of cos t sin t; i.e., y p t a cos t b sin t for some constants a,b. Then y p t ky p t d dt a cos t b sin t k a cos t b sin t b ka cos t bk a sin t cos t. Equating coefficients of cos t sin t on the two sides of this last equation leads to b ka bk a 0 a k k 2 2, b k 2 2. This agrees with the result obtained by variation of parameters. In some situations, it is necessary to express the particular solution in a form that is consistent with the forcing term f t cos t. Note that y p t. k cos t sin t k 2 2. k cos t. k 2 2 k 2 2 k 2 2 sin t. 6
7 Let. k. k 2 2 k 2 2 note further that 2 2 k 2 k k Since the sum of 2 2 equals one, it follows that we can always find an angle such that cos, sin ; i.e. cos sin Tan or Tan 1. Now we can write y p t.. k cos t. k 2 2 k 2 2 k 2 2 sin t. k 2 2.cos cos t sin sin t k 2 2 cos t, from which it is evident that y p t is a periodic function having the same period as the forcing term, but with a phase shift equal to. In addition it is clear that the amplitudes of the forcing term the particular solution are equal to respectively. k 2 2 The following plot shows f t y p t plotted on the same axis from which the delay in y p t by the amount is apparent, amplitude f(t) 1, amplitude of Yp.5 7
8 . 8
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