TS EAMCET 2016 Code A Solved Paper

The given differential equation is,
y′=‌

1

e−y−x

‌

dy

dx

=‌

1

e−y−x

Write the above equation as,
‌

dx

dy

=e−y−x
‌

dx

dy

+x=e−y
Compare with ‌

dx

dy

+Px=Q
So,
P=1,Q=e−y
Therefore,
$$ I F \;\; =e^{∫ P d y} $
‌‌=e∫1dy
‌‌=ey$
The solution of the given differential equation is,
x⋅ey‌‌=∫ey⋅e−ydy+C
x⋅ey‌‌=∫1dy+C
x‌‌=e−y(y+C)