If y^1/5+y^-1/5=2x and (x^2-1)d^2ydx^2+ xdydx+ky=0, then +k is equal to

Question

If $y^{1/5}+y^{-1/5}=2x$ and $(x^2-1)\dfrac{d^2y}{dx^2}+\lambda x\dfrac{dy}{dx}+ky=0$, then $\lambda+k$ is equal to

Accepted Answer

$y^{1/5}-y^{-1/5}=\pm2\sqrt{x^2-1}$. Then $y'(y^{1/5}-y^{-1/5})=10y\Rightarrow y'\cdot2\sqrt{x^2-1}=10y\Rightarrow y'\sqrt{x^2-1}=5y$.

Differentiating: $y''\sqrt{x^2-1}+y'\dfrac{x}{\sqrt{x^2-1}}=5y'\Rightarrow y''(x^2-1)+xy'=5y'\sqrt{x^2-1}=25y$.

So $y''(x^2-1)+xy'-25y=0\Rightarrow\lambda=1,\ k=-25\Rightarrow\lambda+k=-24$.

Solutions · 1