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```        dy/dx=1+x+y+xy
y(-1)=0
then what is the value of funtion y ?
```
8 years ago 419 Points
```							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

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```
8 years ago
```							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
```
8 years ago
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