JEE Main 26 Feb 2021 Shift 1 Solved Paper

Given, y2=a[x+‌

√a

2

],a>0 ...(i)
Differentiating both sides w.r.t. ' x ',
2y‌

dy

dx

=a[1+0]=a . . . (ii)
Use Eq. (ii) in Eq. (i) to eliminate the constant ' a '.
y2=2y‌

dy

dx

(x+√2y√‌

dy

dx

)
y2−2xy‌

dy

dx

=2√2⋅y√y⋅‌

dy

dx

√‌

dy

dx

Squaring on both sides,
y4+4x2y2(‌

dy

dx

)2−4xy3‌

dy

dx

=8y3(‌

dy

dx

)3
Thus, degree of above differential equation is 3 and its order is 1 .
Difference between degree and order =3−1=2