JEE PYQ: Differential Equation Question 27
Question 27 - 2020 (07 Jan Shift 1)
Let $x^k + y^k = a^k$, $(a, k > 0)$ and $\frac{dy}{dx} + \left(\frac{y}{x}\right)^{1/3} = 0$, then $k$ is:
(1) $\frac{3}{2}$
(2) $\frac{4}{3}$
(3) $\frac{2}{3}$
(4) $\frac{1}{3}$
Show Answer
Answer: (3)
Solution
$kx^{k-1} + ky^{k-1}\frac{dy}{dx} = 0$, so $\frac{dy}{dx} = -\left(\frac{x}{y}\right)^{k-1}$. Comparing with $\frac{dy}{dx} = -\left(\frac{y}{x}\right)^{1/3}$: $\left(\frac{x}{y}\right)^{k-1} = \left(\frac{y}{x}\right)^{1/3} = \left(\frac{x}{y}\right)^{-1/3}$. So $k-1 = -\frac{1}{3}$, $k = \frac{2}{3}$.
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