TS EAMCET 11-Sep-2020 Shift 2 Solved Paper

Given, differential equation
xdy−ydx=√x2+y2dx...(i)
Putting, y=vx and differentiating w.r.t. x, we get
dy=vdx+xdv...(ii)
From Eq. (ii), Integrating on both sides, we get
ln(v+√1+v2)=ln‌x+C
Putting v=‌

y

x

, we get
ln(‌

y

x

+‌

1

x

√x2+y2)=ln‌x+C
ln{‌

1

x

(y+√x2+y2)}−ln‌x=C
ln{‌

y+√x2+y2

x2

}=C...(iii)
[∵ln‌a−ln‌b=ln‌

a

b

]
Given that, y=1,x=√3
ln‌l=C or C=0‌‌[∵ln‌1=0]
Putting the value of C in Eq. (iii), we get
ln{‌

y+√x2+y2

x2

}=0
or ‌

y+√x2+y2

x2

=e0=1
y+√x2+y2=x2 or x2−y=√x2+y2
Squaring on both sides, we get
(x2−y)2=x2+y2