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Question
Find the particular solution of the differential equation
(1 – y2) (1 + log x) dx + 2xy dy = 0, given that y = 0 when x = 1.
Solution
Given:
(1−y2)(1+logx)dx+2xydy=0
⇒(1−y2)(1+logx)dx=−2xydy
`=>((1+logx)/(2x))dx=-(y/(1-y^2))dy" ......(1)"`
Let:
1+logx=t and
(1−y2)=p
`=>1/xdx=dt " and " −2ydy=dp`
Therefore, (1) becomes
`intt/2dt=int1/(2p)dp`
`=>t^2/4=logp/2+C "......(2)"`
Substituting the values of t and p in (2), we get
`((1+logx^2))/4=log(1-y^2)/2+C " ......3"`
At x=1 and y=0, (3) becomes
`C= 1/4`
Substituting the value of C in (3), we get
`(1+logx^2)/4=log(1-y^2)/2+1/4`
⇒(1+logx2)=2log(1−y2)+1
Or
(logx2)+logx2=log(1−y2)2
It is the required particular solution
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