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 $\mathrm{k}$ is :
Correct Option: , 3
Solution:
$x^{k}+y^{k}=a^{k}(a, k>0)$
$k x^{k-1}+k y^{k-1} \frac{d y}{d x}=0$
$\frac{\mathrm{dy}}{\mathrm{dx}}+\left(\frac{\mathrm{x}}{\mathrm{y}}\right)^{\mathrm{k}-1}=0 \Rightarrow \mathrm{k}-1=-\frac{1}{3} \Rightarrow \mathrm{k}=2 / 3$