Question

Question 1 3 pts The solution of the Initial-Value Problem (IVP) Į (x + y)dx – xdy = 0 1 y(1) = 0 is given by Oy= (x + y) In x None of them Oy= xel-1-1 O y = x ln(x + y) Oy= x In x

Answer 1

Given da! = Rpx-xp(htx) o=hpr-xphtepx <= => xdx = aph-apr re se dre ndy-y du x? ap d(2) xdy-ydu = -> dxud() sal da x? =xar te; ,cis coristont and s dhe me =lnx sanan so falt) = Inx-c so, 3 => y = je (lnx-e) Now, y(a)=0 :) when se=1; y=0 using this we gel From n

0 = 1 (In 1-c) => 0 = 110-c) [ en 1=o => o =0-e get => c=0 Hence from ③ we y = x (lnx-o) lar. => y=x Last option is conpeet.
Remember the formula i used. In a line during handwritten may be little problem that is

y/x = ln x +C.

.If you have any problem please comment. I will explain