JEE PYQ: Differential Equation Question 11
Question 11 - 2021 (25 Feb Shift 1)
If a curve passes through the origin and the slope of the tangent to it at any point $(x, y)$ is $\frac{x^2 - 4x + y + 8}{x - 2}$, then this curve also passes through the point:
(1) $(4, 5)$
(2) $(5, 4)$
(3) $(4, 4)$
(4) $(5, 5)$
Show Answer
Answer: (4)
Solution
$\frac{dy}{dx} = (x-2) + \frac{y+4}{x-2}$. Let $X = x-2$, $U = y+4$: $\frac{dU}{dX} = X + \frac{U}{X}$. I.F. $= \frac{1}{X}$. Solution: $\frac{U}{X} = X + c$, i.e. $\frac{y+4}{x-2} = (x-2) + c$. Through $(0,0)$: $\frac{4}{-2} = 4 + c$, so $c = -6$. Check $(5,5)$: $\frac{9}{3} = 9 - 6 = 3$. Yes, it works.
BETA
