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Question:

Find the particular solution of the differential equation dydx=1+x+y+xy, given that y=0, when x=1.

Solution:

dydx=(1+x)(1+y)⇒∫dy(1+y)=∫(1+x)dx⇒log|1+y|=x+x22+C(i)Putting x=1 and y=0 in (i), we getC=−12.∴log|1+y|=x+x22−12.