Solutions to Midterm II. of the following equation consistent with the boundary condition stated u. y u x y

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1 Sltis t Midterm II Prblem : (pts) Fid the mst geeral slti ( f the fllwig eqati csistet with the bdary cditi stated y 3 y the lie y () Slti : Sice the system () is liear the slti is give as a sperpsiti f a hmges ad a particlar slti y y y We start by cmptig the hmges slti; H P Characteristic Eqati: Characteristic Crves: Hmges slti: dy d C y y H ( f y Net we slve fr a particlar slti t () We assme P y a by If we sbstitte the particlar slti it () We btai ay b 3 a b 3 Hece Geeral slti: O y : Hece y y P 3 y y ( f 3 ( ) f ( y 3y

2 Prblem : (3pts) Slve fr ( 6 5 y y Sbject t ad 4 6 y bth the lie y Slti : Characteristic eqatis: Characteristic Crves: Geeral slti: Bdary cditis: dy d 5 5 4(6)() 5 (6) 3y C ad y C y f 3y g y f g ad y 3 f g 4 6 () ifferetiate () f g 3 f g 4 6 (34) We slve eqatis (3) ad (4) simltaesly Itegrate (56) f S that Hece Simplify Bdary cditi S that g 6 f 6 (56) k 3 g k 3 3y k 3 3y g y k y 3 y f y f 3y g y k 3 3y k y 3 y y 4y 6y 5y k k k k k y 4y 6y 5y k

3 Prblem 3: (5pts) Shw that there is slti f f i g I three dimesis less Hits: Use the divergece therem () fdv gds (3) FdV F ds ˆ Slti 3: We itegrate () fdv dv (4) We w apply the divergece therem the right side f (4) fdv dv ds ˆ ds gds Q.E. Prblem 4: (45pts) A thi sheet f metal cicides with a it sqare i the y plae. Iitiall the temperatre i the sheet is h y. If there are srces f heat i the sheet fid the temperatre at ay pit at ay sbseqet time give that the right ad left faces f the sheet are islated the temperatre at the lwer edge is maitaied at zer ad the temperatre at the pper edge is t cs. prescribed by Slti 4: he gverig eqati t be slved is t y y t t t t cs h y () We defie r slti as the sm f a steady ad a trasiet cmpet t y ( t () We w sbstitte () it () ad rearrage )

4 t y y y y ( y ( ( ( cs y ( ) h( (3) We w make the fllwig chices fr t t y y t t t t g y (4a) Ad fr y y y y y cs (4b) We slve (4a) by assmig separable sltis f the frm We w sbstitte (5) it (4a) t X ( ) ( ( (5) X X X We w divide thrgh by t X ( ) ( ( X X (6) he left side f (6) is a fcti f t ly. he first term the right is a fcti f ly ad the last term a fcti f y ly. he ly way fr bth sides t be eqal is if they are each eqal t sme cstat say. X X By ispecti we have chse the cstats t be egative. S that X ( ) X ( ) ( ( t t (7a)

5 Fr -trivial sltis we select r bdary cditis as We w slve (7) () () X ( ) X (7b) c cs c si c csy c siy X ( ) t c t ( 3 4 We w apply the hmges bdary cditis ( ) c X X ( ) c si Fr trivial sltis c... S that X ) c cs Fr trivial sltis 5 ep ( (8) ( ) c3 ( ) c4 si m m m... S that ( b simy Hece t a ep m t m m m m he sltis m simy ep m t ( cs m Sice the system is liear a cmplete slti is give as a liear cmbiati f sltis si my ep m t m cs m ( he cstats A m are give as A m my cs g( si ddy We w lk at sltis t (4b). We assme separable sltis f the frm y GH y s that G ( ) G( ) ad H ( H( (9ab) () X ( ) X

6 We already slved (9a). he sltis are give i (8). S we lk fr the sltis f (9b) ( c3 ep 4 y c ep y ) c c ( 3 4 he sltis y ( b sih y... sihy ( y cs... Sice the system is liear a cmplete slti is give as a liear cmbiati f sltis ( ( y cs At this pit we apply the hmges bdary cditi sih y ( ) cs cs sih By cmparig cefficiets the tw sides f the abve eqati we see that ly the term srvives yieldig sih S that ( cs sih sih y Fiall t cs sih y m simy cs ep m sih m t

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