Question

Thank you.

1. Find all solutions of each of the following homogeneous boundary value problems. (a) y"+y=0, y(0) = 0, y'(T) = 0. (b) y" – 2y' + 2y = 0, y(0) = 0, y(T) = 0. (c) y + y = 0, y(0) = 0, y(L) = 0, where L is a positive real number.

Answer 1

- - The giver differential equation is yu+y=0 set y= Amre be the trial solution The auniliary equs is mrl=0 rima i zotin The solution of the equation is yw = e . {Acost.x+B sinin} > fos X+ Bsinko and y'(X) = - Asina + B cosx () using the initial condition 40) 20 in equa (i) O= A. Caso + B sino 27 0= A. I TB.0 27 Azo also using y' (T) =0 in () weget o= - Asinta B GOST o = 0 + B.(-4) 7 B -0 .. The general solution is gouzo

by The given differential equation is y'-24'tayo det tema be the trial solution of the above doff equs The auniliary equation is nr_2m +2 =0 ima 2 - 5 4 - 4.1.2 = 2 1 2 2 ... The general solution of the above diff.eque is anga er { fursa Absix} using the initial condi Hon yo) 20 we get ose { Acos o + Bain of >) OD! {A.178.0} 2) Azo and port using 4920 we get, o e do. cost + B sinto } >ose". Bio 7 B is arbitarry The general solution is yn) = Bersinx The diff.equis is yrty .... Let uzten be the trich sold of the above toffiequ? The anniliary eque is mollao mati Гtu am та solution 7) - 4 сотт - sinx using the initial condition yo) o we get 10) = Aaso & B simo ey also ustry 414 20 we get y (4 = AlosL & B sinh → 0 = 0 Cool+ B.Sine B SimL 20 , where his positive integer then B20 Then B is arbitary if L&NT or L=nt

The general solution is ya so 1 B sink if Lint if L=nt