JEE PYQ: Differential Equation Question 6
Question 6 - 2021 (18 Mar Shift 1)
The differential equation satisfied by the system of parabolas $y^2 = 4a(x+a)$ is:
(1) $y\left(\frac{dy}{dx}\right)^2 - 2x\left(\frac{dy}{dx}\right) - y = 0$
(2) $y\left(\frac{dy}{dx}\right)^2 - 2x\left(\frac{dy}{dx}\right) + y = 0$
(3) $y\left(\frac{dy}{dx}\right)^2 + 2x\left(\frac{dy}{dx}\right) - y = 0$
(4) $y\left(\frac{dy}{dx}\right)^2 + 2x\left(\frac{dy}{dx}\right) - y = 0$
Show Answer
Answer: (3)
Solution
From $y^2 = 4a(x+a)$: $2yy’ = 4a$, so $a = \frac{yy’}{2}$. Substituting back: $y^2 = 4\cdot\frac{yy’}{2}\left(x + \frac{yy’}{2}\right) = 2yy’x + y^2(y’)^2$. So $y(y’)^2 + 2xy’ - y = 0$.
BETA
