Let x^k+y^k=a^k, (a,k0) and dydx+( yx)^1/3=0, then k is:

Question

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

Accepted Answer

Differentiating $x^k+y^k=a^k$: $kx^{k-1}+ky^{k-1}\dfrac{dy}{dx}=0\Rightarrow\dfrac{dy}{dx}=-\left(\dfrac xy\right)^{k-1}$.

Comparing with $\dfrac{dy}{dx}=-\left(\dfrac yx\right)^{1/3}=-\left(\dfrac xy\right)^{-1/3}$: $k-1=-\dfrac13\Rightarrow k=\dfrac23$.

Solutions · 1