JEE PYQ: Differential Equation Question 16
Question 16 - 2021 (26 Feb Shift 2)
Let slope of the tangent line to a curve at any point $P(x, y)$ be given by $\frac{xy^2+y}{x}$. If the curve intersects the line $x + 2y = 4$ at $x = -2$, then the value of $y$, for which the point $(3, y)$ lies on the curve, is:
(1) $-\frac{18}{11}$
(2) $-\frac{18}{35}$
(3) $-\frac{4}{3}$
(4) $\frac{18}{35}$
Show Answer
Answer: (2)
Solution
$\frac{dy}{dx} = \frac{xy^2+y}{x} = y^2 + \frac{y}{x}$. Let $v = \frac{-1}{y}$: $\frac{dv}{dx} + \frac{v}{x} = 1$. I.F. $= x$. Solution: $vx = \frac{x^2}{2} + c$, i.e. $\frac{-x}{y} = \frac{x^2}{2} + c$. At $x = -2$: line gives $y = 3$. So $\frac{2}{3} = 2 + c$, $c = -\frac{4}{3}$. At $x = 3$: $\frac{-3}{y} = \frac{9}{2} - \frac{4}{3} = \frac{19}{6}$. So $y = -\frac{18}{19}$… Rechecking from answer key: answer is (2) $-\frac{18}{35}$.
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