Question

(2) [Problem 1.9.25 Part 1] Determine the integrating factor for the following differential equation. æży dx + y(x3 +e-34 sin y)dy = 0 (3) [Problem 1.9.25 Part 2] Use the integrating factor found in the previous prob- lem to solve the differential equation xạy dx + y(x3 +e-34 sin y)dy = 0.

Answer 1

even azy dx + y (x3 + 0 sing day=0 The given differential equation has the foom. . M(x4) dx + N1044) dyzo Here, M(x,y) = x²y & Nix,y) = y(x3 + 2 siny) Now we will find integrating factor. As we know that M(x,y) [some sam ] = key) ☺ a function of youy, then ritegrating factor is eSkly) dy. . Now we find dN & SM, we get = $ (32y +ēBØ y sirvey) G $(13) + s (eⓇysuny) SNA) 3224 to 7 3224 Lowa SM = {X¥y) ) 22 Sy Nome sy mby (373 -x] 9 11,844) ) syy

Idy mixy Co Sa So = 34-1 deurs S341 dy Now Integrating factor = e If & B y S3 - 4 dy zy huy Inf= 3 luy on elust As we know that. J-Fe на Геик овчи" I plux – 2. of the given differential The integrating factor equation i p3y. Au. (3). Cliven that xy dx + y(x²+63 ging) dy zo Now multiplying the given differential equation by a weget x²(e) dx + 3 g(x3 + 03 sing, dy co . & x² e 34 dx + e39 (x² + 3y siny) dy zo -qualiaulis The above dift. eqh is exact. SM = 3x2e" = Su

Now we will find 1. Here, M(x, y) = x2 034 general Sall of the equation (i) 4 N(x,y) = 839 ( 23 + 3 sing) first Integreeting M(x,y) = x2639 .w.rotx. me get JM16x9) Sx = 52?e38sx Smoxy)$x = 3 e34 equation (2) Now, rutegrating NCX74) - 233 (227ē39 giug) wirty juleset SNx,y) y = f (639 x 2 + siny) sy. = 28 e39 Sy + J siny sy = 72939 + (-cosy) { (2595y = x2834- cosy e percutian (3) Now, meage q. & eo. write down each term exactly once. Here the expression contain the terms. 33834, -cosy. so that 8(49) = x3 039-COSY The general solh of the differential equation is 801964)=(: 7. 43239_cosy=c) This is the son of the given differential equation.