Question

Solve:

$u_t-u_x_x=e^{-t}, x\epsilon R, t>0 \\ u(x,0)=e^{-x}, x\epsilon R$

Answer 1

rae wh Du ut at Civen PDE- ut - 4 xx = tet (0 - Di?) u = text - where ay - Du Mere D-D's is not a linear Uxx 2 = D²u factor in D and D' (ciat Let z=&A het + kik be a trial solution of the homogeneous ean of given P.D.E. ie. (0-0²2,420 then Dz= 22 - Z shie hit this kie 2x2 homogeneous Putting these values in the goon differential eqh, un ged {{ Arhieht + tx_ Aile bit + cx} = 0 - Ail hin ke?) e hit ti kix - o himle?? = 0 as his lei? e hit + lexe Theis replacing hi by lee, the most general solation of the homogeneous PDE ý le complementersy factor of 6 is - - Cof. = 8. Akt + lege where A and ki are arbitrary constants.

Particular integral Pisaés D-D 2 Elttoa sina att be IF (DD) at & be (-1) frantong given blince general solepsion of differential equation. ext) = Cofit P.I. 1469%, +) 2 Faieleto + 1642 449,0) = € *=> I Azelex-1=ė {patting t=0) - Aiekix = It e-x => Ao e Rox + A, ek x + Azekaz = itex compairing the terms of both sides we get AO = 1, Ko= 0,A, 1, k and all other Aiard kils are zero 1e Aiako wlin i flince solution of given P.D.ET (by ean ula, 6) = so ekolttkok ta 12art kit + A e katt kat A, ettet 12 + - + lelottlox (using -t 6 ) a lucent) = l tettx

L.I. so that Theorem. To show that F(D, D') eax+by = F(a,b) ear+by Proof. We know that F(D, D') consists of terms of the form C D'D's Also D' (ear+by) = d'ear+by and D's (eax+by) = b eax+by (Cs D'D') (eartby) = Cipa' b' eax+by The theorem follows by combining the terms of the operator F(D, D'). We now discuss method of finding C.F. of (1). Consider F(D, D') z=0 .. (2) From the above theorem we see that chr+ky is a solution of (2) provided F(h,k) = 0, so that .... (3) z = A eh; x+k;y in which A,, h,, k; are all constants, is also a solution provided that h, k, are connected by the relation Fh; ,k;)=0 e ... (4)