+Px=Q, where P and Q are functions of y or constant terms. Here, P=−1 and Q=y=1 ∴ IF=e∫Pdy=e∫(−1)dy=e−y Now, general solution is given by x.IF=∫IF.Qdy+C1 ⇒ x.e−y=∫en−y(y+1)dy+C1 ⇒xe−y=(y+1)(
e−y
−1
)+∫1.e−ydy+C1 ⇒xe−y=−e−y(y+1)−e−y+C1 ⇒x=−(y+1)−1+Cey[ondividingbye−y] ⇒x+y+2=Cey On taking log both sides, we get log (x+y+2)=logC+1ey ⇒ log (x+y+2)=logC1+logey [∵logmn=logm+logn] [putC=logC1] ⇒log (x+y+2)=C+y which is the required solution.