Question

20.3 Solve the following partial differential equations for u(x, y) with the bound- cary conditions given: du . (a) x + xy = u, u= 2y on the line x = 1; b) 1 + x n = xu, u(x,0) = x.

Answer 1

solution: ② x dateyall, u= ry on the line xel. By tag Lagranges equation are given by . du ] x2 . dy zo syac, = constant Taking 1st & last ratios, we get, u-ex : [yza] " а W x du - udx=-axdi ar x du-uds = ax ax { piriding by a dlu) = -a allogx) Integrating both sides, we get, a = -6 logs & C2 a; a (x, y) = -axloga + C₂ x the .. wex, y) = xy log x + Ca.X. - ③ buveu, uszy , on xal so by ③ 2y = - y logi + C2 272 62 ar asce [ by ® ] a 202 C2 a 2y = u(x,y) + xy logx7 u(x, y) + xy logx = any ulay) = 2xy - xylogx which is required sobition a a

6 buren, tx du z au Mr. I x y = xu- = 0 By zagrange's equation's, a du a dy du O U KU-1: si ha oaxa e a constanta Taking last two ralios, we have, dy i du x xel-1 dy a du sous- — с си — 1 Integrating, we have Guyzz Y = Go T log cao a du z dz + au atat N, Y = log(xU-1) + C2 [ixae] Given that, "ulx, o) zx so by ② oa rog (x:x-1) + Co. a, C2 E-log (x²-1) ar C2 = -log (971) ["xzeil Now putting the values of G & C in the above equation, we have, y log (& U-1) = – log (x2 - 1) iz log (xn-1) - log (x7) elog au- xu e e7 » uz [lt (22) e ] : u(x,y) = 2 [it los 1) e I which is the required XL solution.