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Find the particular solution of the differential equation (1 – y^2) (1 + log x) dx + 2xy dy = 0, given that y = 0 when x = 1. - Mathematics

Find the particular solution of the differential equation

(1 – y2) (1 + log x) dx + 2xy dy = 0, given that y = 0 when x = 1.

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Solution

Given:

(1y2)(1+logx)dx+2xydy=0

(1y2)(1+logx)dx=2xydy

`=>((1+logx)/(2x))dx=-(y/(1-y^2))dy" ......(1)"`

Let:

1+logx=and

(1y2)=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(1y2)+1

Or

(logx2)+logx2=log(1y2)2 

It is the required particular solution

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