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Question
Find the particular solution of the differential equation dy/dx=1 + x + y + xy, given that y = 0 when x = 1.
Solution
`dy/dx=1+x+y+xy`
`dy/dx` = 1 + x + y + xy
`dy/dx = 1( 1 + x ) + y ( 1 + x)`
`dy/dx=(1+x)(1+y)`
`dy/(1+y)=(1+x)dx`
Integrating both sides:
`intdy/(1+y)=int(1+x)dx`
`log|1+y|=x+x^2/2+C`
y = 0 when x = 1 (given)
`log1=1+1/2+C`
`C=−3/2`
`⇒log|1+y|=x+x^2−3/2` is the required solution.
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