dy/dx=1+x+y+xy y(-1)=0 then what is the value of funtion y ?

							
Dear Nikhil
Factorise RHS
RHS=1+x+y(1+x)= (1+x)(1+y)
dy/1+y = (1+x)dx
now integrate both sides
ln(1+y)=x+x2/2 + C
put x=-1 and y=0
0=-1 + 1/2 +C
C=1/2
ln(1+y)=x+x2/2 +1/2
1+y = ex+x2/2 +1/2
y=ex+x2/2 +1/2 -1
 
All the best.                                                           
AKASH GOYAL
AskiitiansExpert-IIT Delhi
 
Please feel free to post as many doubts on our discussion forum as you can. We are all IITians and here to help you in your IIT JEE preparation.
Win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian.
Now you score 5+15 POINTS by uploading your Pic and Downloading the Askiitians Toolbar  respectively :Click here to download the toolbar..
						

							
dy/dx = 1+x+y+xy
dy/dx = 1+x +y(1+x)
dy/dx = (1+x)(1+y)
now using variable separation
dy/(1+y) = (1+x)dx
log(1+y) = (x+x2/2)+c
y(-1) = 0 so
c = 1/2
log(1+y) = x+x2/2+1/2
this is the required solution
approve if u like it