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The curve amongst the family of curves,...

The curve amongst the family of curves, represented by the differential equation `(x^2-y^2)dx+2xydy=0` which passes through (1,1) is

A

(a) a circle with centre on the Y-axis

B

(b) a circle with center on the X-axis

C

(c) an ellipse with major axis along the Y-axis

D

(d) a hyperbola with transverse axis along the X-axis.

Text Solution

Verified by Experts

The correct Answer is:
(b)
Given differential equation is
`(x^(2)-y^(2))dx+2xy dy=0,` which can be written as
`dy/dx=(y^(2)-x^(2))/(2xy)`
Put y=vx [`therefore` it is in homogeneous form]
`dy/dx=v+ x(dv)/dx`
Now, differential equation becomes
`v+x(dv)/dx=(v^(2)x^(2)-x^(2))/(2x(vx)) rArr v+x(du)/dx=((v^(2)-1)x^(2))/(2vx^(2)`
`rArr x(dv)/dx=(v^(2)-1)/(2v)-v = (v^(2)-1-2v^(2))/(2v)`
`rArr x(dv)/dx=(1-v^(2))/(2v) rArr int(2v dv)/(1+v^(2))=-int dx/x`
`rArr ln (1+v^(2))=-ln x - ln C`
`[therefore int (f'(x))/(f(x))dx rArrln abs(f(x))+C]`
`rArr ln abs((1+v^(2))Cx)=0 [therefore ln A+lnB=lnAB]`
`rArr (1+v^(2))Cx =1 [log_(e) x=0 rArr x = e^(0) =1]`
Now, putting `v=y/x,` we get
`(1+y^(2)/x^(2))Cx=1 rArr C(x^(2)+y^(2))=x`
`therefore` The curve passes through (1,1), so
`C(1+1)=1 rArr C=1/2`
Thus, required curve is `x^(2)+Y^(2)-2x=0,` which req\present
A circlc having contrc (1,0)
`therefore` The solution of given differential equation represents
a circle with centre on the X-axis.
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