Question

(y)2 – 2yy" + y2 = 0. Use an exponential ansatz to find two (possibly complex-valued) solutions yi and y2 of the differential equation. Do any of the theorems assure us that cıyı + c2y2 will also be a so- lution of the differential equation? Answer yes or no, either naming the theorem/principle or else briefly explaining why not.

Answer 1

Let fid 4 x हनु मा - l We get TH 21 or wita y = is the solution on substituting yyy"+ya e2 22x - 2012 22x + 24 x = 0 (- e2+1) e 2 rx = 0 For solution e 2 rx to So – 42+1=0 We get de 2 = 1 r=#1 So we get twee solutions y, = et and if get+ca ex is a solution or not just substitute in (y")2-24 y"+y² (a excex) ²2 (Cierta ex) (Gert cze ) + (Getthet ex ge-262- Get 34 -20C2. - 2002-26, C2 -X e Y 2= Lets verify 2 L. Hos » - 4 G C is For solution one of G or C₂ =0 then it yz ay, or y = 242 So Yagy, tc2Y2 is not a solution. means y=ay, +C₂ 42