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438 Note 14 for MATH 251 at PSU

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Date Created: 02/06/15
Differential Equations LECTURE 14 More On The Wronskian Last lecture we introduced the Wronskian of two functions yl and yg meaoj wouoeuowo We saw that if Wy1y2t 7 0 then yl and yg are linearly independant ie the only constants 01 and 02 that satisfy 6124105 6224205 0 are 01 62 0 In other words two functions are linearly independent if they aren t constant multiples of each other We also saw that in the context where yl and y2 are solutions to the linear homogeneous equation pty qty rty 0 Wy1 y2t 7 0 is precisely the condition for the general solution of the differential equation to be 2405 0124105 62242037 ie for yl and yg to be a fundamental set of solutions Thus if yl and yg are a fundamental set of solutions they are linearly independent If we have two solutions yl and yg which are linearly dependent on the other hand then they cannot possibly be a fundamental set of solutions as they have a zero Wronskian There was one point last lecture that we should clear up now We know that if our initial data is at to yl and y2 will be a fundamental set of conditions if and only if Wy1 y2t0 7 0 But this is a condition only at one point What happens if yl and y2 have a nonzero Wronskian only at to but not at nearby points This would be problematic since then our condition would tell us that yl and y2 are a fundamental set of solutions for initial dat at to but not for any initial data near to The possibility of this should seem odd since we know there should be a unique solution on any interval around to where we have continuity So how do we know that this can t happen The answer is something called Abel s Theorem 1 Abel7s Theorem You may notice that throughout our entire discussion of the Wronskian we have yet to actually use the differential equation beyond deriving the formula for the Wronskian assuming that yl and y2 satis ed some differential equation Fortunately when yl and y2 are solutions to a linear homogeneous differential equation we can say a bit more about their Wronskian THEOREM 141 Abel s Theorem Suppose y1t and y2t solue the linear homogeneous equa tion Wt WW 6175 0 where pt and qt are continuous on some interval 11 Then for a lt t lt 1 their Wronskian is giuen by t muwmwmmwm ww where to is in 11 Differential Equations Lecture 14 More On The Wronskian lf Wy1 y2t0 7 0 at some point to in the interval a b then Abel s Theorem tells us that the Wronskian can t be zero for any t in a 1 since exponentials are never zero This assures us that we can change our initial data without crossing points of discontinuity of the coef cient functions without worry that our general solution will change Another advantage of Abel s Theorem is that it lets us compute the general form of the Wron skian of any two solutions to the differential equation without knowing them explicitly This is useful for example with regard to reduction of order where we only begin by knowing a single solution The formulation given in the statement of the theorem isn t so computationally use ful however because we might not have a precise to in mind let alone knowing the value of the Wronskian there But if we apply the Fundamental Theorem of Calculus things simplify nicely Wy1y2t Wy1y2t05 fto 17z dz ceifptdt What is this constant 0 Well it doesn t really end up mattering If we know the value of the Wronskian at one point we can compute it but our general interest in the Wronskian mostly involves knowing its general form As long as we know 0 7 O that s all that matters to us EXAMPLE 141 Compute up to a constant the Wronskz39an of two solutions yl and yg of the di erentz39al equation t4yn 7 2t3y 7 tsy 039 First we need to put the equation in the form speci ed in Abel s Theorem We do this by dividing by the leading coef cient 2 gm 7 2y 7 t4y 039 So Abel s Theorem tells us W ceifigdt 36th ctz E 0k greatbut the main virtue of this is that it gives us a second way to compute the Wronskian A general rule in mathematics is that whenever you can compute something in two different ways something good will happen In this case we know by Abel s Theorem that Wy17 y2t ceifmtwt39 On the other hand by de nition u 2 Wm y2gtlttgt 3 yg t y1lttgtyglttgt 7 WW 9 u Setting these equal if we know one solution yl t we re left with a rst order differential equation for yg that we can then solve Let s see this with an example of reduction of order we did the traditional way EXAMPLE 142 Suppose we want to nd the general solution to 2t2y ty 7 3y 0 and we re given that y1t t 1 is a solution We need to nd a second solution that will form a fundamental set of solutions with yl Let s compute the Wronskian both ways 1 CH W WW y2gtlttgt ygrl m2 y t 1 ygt z cei lnm ct Differential Equations Lecture 14 More On The Wronskian This is a rst order linear equation with integrating factor Mt efrl dt 511 25 Thus my ct t i 2 t 92 7 EC 2 y2t get kt l Now7 we can choose constants c and k Notice that k is the coefficient of 25 17 which is just y1t So we don t have to worry about that term7 and we can take k 0 We can similarly take 0 g and so we ll get y2t 25 which is precisely what we had gotten when we did reduction of order the traditional way D

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