Question

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

• A. $\frac{2}{3}$
• B. $\frac{1}{3}$
• C. $\frac{3}{2}$
• D. $\frac{4}{3}$

Answer 1

The correct option is A $\frac{2}{3}$

$x^k+y^k=a^k\Rightarrow k{x}^{k-1}+k{y}^{k-1}\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=-{\left(\frac{x}{y}\right)}^{k-1}\Rightarrow \frac{dy}{dx}+{\left(\frac{y}{x}\right)}^{1-k}=0\Rightarrow 1-k=\frac{1}{3}\Rightarrow k=\frac{2}{3}$