Question

In Problems 1-3, solve the given DE or IVP (Initial-Value Problem). [First, you need to determine what type of DE it is. 1. (2xy + cos y) dx + (x2 – x sin y – 2y) dy = 0. 1 dy 2. + cos2 - 2.cy y(y + sin x), y(0) = 1. + y2 dc 3. [2xy cos (2²y) – sin x) dx + x2 cos (x²y) dy = 0. (1+y! x" y® is an 4. Determine the values of the constants r and s such that M(x,y) Integrating Factor for the following DE. (2y2 + 4x²y)dx + (4xy + 3x3)dy = 0.

solution for all 4 please

Answer 1

0 The given DE is (2ay + Cosy) dae + (x²-xesiny -2y) dy zo or aay dat cosy die ta dyan sing dy-zydy = 0 (2xy da tx?dy) + (cosy dx - x sing da) -24 dy so or d(x)) + d(acosy) -zy dy = 0 eo Integrating we get or, @ Jldcapy) + d(xcosy) –2ydy) = 0 oo, Jacary) + Jacacosy) - Jay dy=0 n'y & acosy-2x - 2x -CEO Or, x'y tacosy-y2=c. constant where cis arbitary The giren DE is (ingat y (It sinx) (htye + cosa "2xy dy = y Cyrsinx) dx + cosxdy - 2xy dy a y dxty sinx da Hy2 Or, dy +(cosa dyry sinx dx) = y dx tixydy HY2 dy d Cy cosx) d (ay). Hy2 teosu -2xy dy dx or or dy 07, t

لاله D) 07, Where e is 3 Integrating we get s 14 y + d (y cosx)] = J denyny J ao e + J d (y cosx) of dezyn fan'(X) + y cost a ny²e é arbitary constant. The given DE is [2xy cos(x+y) - Sinx]dx +* cos(x) dy = 09, 2xy cos cxy) dx - sinx dx + x eos (ott) dy= 0 or Exy cos (a) dx + x cos(x'j) dy - sinxdx = 0 Integrating we get J [2xy cos cxta) dx + fx!cos (23 dy] - dc cos Caty)) - / sinada ao Sinx dx=0 og R or coscan cosnad where dis arbitazy Constant.

4 The given DE is (24²+4x2) da +(4xy 1323) dy ao 3 is an integrating dret Measy) = x y factor of the DED We multiplied by M (ny) such DR ® is exact DE that the a 73 (2777472) dx + (4xy+3x3) dy] =0. S12 ( 2 07 42272 5+1) dx + (4X7 lyst 13 28 5) dy = 2 xa p Stl det Ma = 2x²y st?, 48212 yst) N (41 rel yt Bys) ON For exact, ron xr1 3 x 243 N oma 2 roy ie 20%. (942) yst1 48 re? (ST)YS = 4 (841) х ystl + 3 (2+3) X7 1 2 3 2272 ys il 2 (312) aryst + 4 (5+1) 2012 js = 4 (871) 294571 / 3 (2+3) ? Compais both side We get 2 (512) = 4 (rel) 4(5+1) 3 (713) 5+2= 2242 45+42320g .2 Multiplying ♡ by 4 and Substract from @ get 4(812) - (45+4) 24(2012)-3719) we

or 4 58-1 58-5 ral From 2 S=2812-2 228 = 2x122. DB is exact for Therefore the the value of ral and and s=2.