Question

2 + COS- 2.ry dy d 1+y2 = y(y + sin x), 7(0) = 1. 3. [2cy cos(x+y) - sin x) dx + x2 cos (+²y) dy = 0. 4. Determine the values of the constants r and s such that (x,y) = x'y is an Integrating Factor for the following DE. (2y + 4x^y)dr + (4.6y +32)dy = 0.

Answer 1

2. It can be vositten as lity? +(099 – 244 dy - 914+31n a).da lysty on a) de-lity? +6094-24) dy of the fom Mdet Ndy o where M=g² ty sina and N = -1 cosa + 2you 1+ y2 Check bether it's exact or not. DM sy ON Sina + 2y 2y + sina Thus ON OM sy 021 the giver differential equation is exact,

To solve it, first integrate M o.s.t. a keeping y constant ay² – y cosa Integrate w.rot. g only those teams of N which donot contain a -tantly) Solution Eg. (15 & Eq. (2) - C that is, ay²-y cosa-tanlop = c. Plug the initial condition y loot to had the constant c. -1-tan' (n=c C=-1-1/4 Thus the solution becomes ay²-ycosa - tant (y) = -1-1/4

3. It's also of the hom Mda andy where DM an Da M-Zay cos (a²y) -sina and, N = x cos (x²) To check whether it is exact. 2al cos(ay) - ysin lay) 22)- O - 2 cos(ay)-2234 sio (ay) - 29 Cos(?) - ofcos si 24 - 2 cos (a*y) - 2289 sin lary) Thus, DM sy given differerskal equation is enact, To knd the solution, hast integrate Mw.rot. a keeping a constant. that is day cos (2²9) - sina da DN da the

Raycos (y da - Is sina dai S SIDIU Consider say cos(x²4 de da Substhite U = 224 du - day Thus the integral becomes cos (u) du Substitute back the value of u 31 (x²y) :) [2ay coslay) - sma] da Sintary) & Cosa Now integrate w.o.t. terms of N wbich donot contain a. There's term which only contams g. So the Solution is sin (ay) & Cosa-c SM (²4) = C- Cosa y only those DO (hom (o

the Hlay)=xge (4 8th 9 st att 3x8+3 3 a'y arcsin (c-coon) ya – are gin (c-cosa) 4. Multiply the differential equation by Integrating factor (2010992 4224240+) do + '+370 +3 ys) dy Novo M = 22 y 3+2 +49 872 y St After Bulkiplying with integrating factor, the differenhal eg. must be exact, DM ON by Find OM -2(+2) aºy + 4 (SDX sy that is, ost 0723

ON = 460+Day **y$+1 + 310+5)+ 09 Stl 6+2 9 Equating the values 2(3+2) 204 + 4 (3+1 a 418+1) ay 3+1 + 3 (0+3) Xota ys Comparing the coefficients of same temos 210+2) – 4 (0+1) 3-2020 -3) 413 +1 -3 (+3 43-30=5 - (4) Solving (3) and (4) (4) - 4x(3) 58=5 8=1 From (10, S-2 Неосе S=1, 3-2