Question

Зрт Question 1 f (x + y)da - ady=0 The solution of the Initial-Value Problem (IVP) 1 y(1) = 0 is given by Oy= (x + y) In a Oy = x In a Oy= « ln(x + y) 3 = teº-1 None of them n Question 2 3 pts The general solution of the first order non-homogeneous linear differential equation with variable dy coefficients (a +1) + xy = e > -1 equals da 3 Oy= e-* [C(x2 - 1) + 1]. where is an arbitrary constant. Oy=et (C(x2 - 1) + 1), where C is an arbitrary constant. O None of them Oy= e-*[C(x + 1) - 1), where is an arbitrary constant. Oy= ev [C(x - 1) + 1], where is an arbitrary constant Question 3 2 to

Answer 1

team A da Linear differential ĽXCLDE) DE of the form dy + Prz) y =015) ins called LOE. It's integrating factor espo (x) da (IF)

& solution is ginen by (IF) = S (IF) 0143 da te + que (aty , da x dyzo -) (a + y ) d2 = xdy x+y dx =) dy - =) d bil - I + dx IN Y all ďx It is linear DE comparing with egr 0 we get prx) = -1 bayo X It's integrating factor IF = es da -lna = e | 1F='

Solution is given by y lano) Sla") in) dx + c a) in atc. 2 a Applying Ic yoos=0 Eq" @ =) 9 =lningte =) C=0 Eqh @ =) Y ln x =y=x lnx One (2 (@+i) d'Y fay=e da x B =) dyt da y at ata It is LDE. comparing with we eqm get. 2 play Q18) atl Xtr

(IF) x+1 e att - 1 da a 2+1 ESC-sto) dz x-encato - e ex -en rato e If ex (x+ogy Solution is given by y el (x+1) s é (stesse date at 2 yee Sixteje edxto Det! =) yox tc 1 at Het! =) yee = 1 + c (2+1) sajos et [ 26840)--))