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:
Correct Option: , 3
Solution:
$k \cdot x^{k-1}+k \cdot y^{k-1} \frac{d y}{d x}=0$
$\Rightarrow \frac{d y}{d x}=-\left(\frac{x}{y}\right)^{k-1}$
$\Rightarrow \frac{d y}{d x}+\left(\frac{x}{y}\right)^{k-1}=0$
$\Rightarrow \quad k-1=-\frac{1}{3}$
$\Rightarrow \quad k=1-\frac{1}{3}=\frac{2}{3}$