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.

Answer 1

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