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Question

Find particular solution of different equation dydx=1+y21+x2 given that x = 0 y = 1

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Solution

dydx=1+y21+x2
dy1+y2=dx1+x2
tan1y+C1=tan1x+C2
tan1xtan1y+C=0.....(1)
Given that:- x=0,y=1
tan1(0)tan1(1)+C=0
C=π4
Substituting the value of x in equation (1), we get
tan1xtan1y+π4=0
Hence the particular solution of the given differential equation is tan1xtan1y+π4=0.

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