Let Xk

Question:

Let $x^{k}+y^{k}=a^{k},(a, k>0)$ and $\frac{d y}{d x}+\left(\frac{y}{x}\right)^{\frac{1}{3}}=0$, then $k$ is:

  1. (1) $\frac{3}{2}$

  2. (2) $\frac{4}{3}$

  3. (3) $\frac{2}{3}$

  4. (4) $\frac{1}{3}$


Correct Option: , 3

Solution:

$k \cdot x^{k-1}+k \cdot y^{k-1} \frac{d y}{d x}=0$

$\Rightarrow \frac{d y}{d x}=-\left(\frac{x}{y}\right)^{k-1}$

$\Rightarrow \frac{d y}{d x}+\left(\frac{x}{y}\right)^{k-1}=0$

$\Rightarrow \quad k-1=-\frac{1}{3}$

$\Rightarrow \quad k=1-\frac{1}{3}=\frac{2}{3}$

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