BITSAT 2019 Paper

(c) We have y - x

dy

dx

= a (y2+

dy

dx

)
⇒ ydx - xdy = ay‌2dx + ady
⇒ y(1 - ay)dx = (x + a)dy
⇒

dx

x+a

−

dy

y(1−ay)

= 0
Integrating, we get
log(x + a) - log y + log(1 - ay) = log C
or log

(a+x)(1−ay)

y

= logC i.e. (x + a)(1 - ay) = Cy
Since the curve passes through (a,−

1

a

)
∴ 2a x (1 + 1) = −

C

a

i.e C = -4a‌2
So, (x + a)(1 - ay) = - 4a‌2y