Q.$2y=\sqrt{x+1}+\sqrt{x-1}$, show that $4(x^2-1)y_2+4xy_1-y=0$

Given $2y=\sqrt{x+1}+\sqrt{x-1}$. Let $u=\sqrt{x+1}$, $v=\sqrt{x-1}$, so $2y=u+v$.

Step 1 — differentiate: $2y_1=\dfrac{1}{2\sqrt{x+1}}+\dfrac{1}{2\sqrt{x-1}}=\dfrac{1}{2u}+\dfrac{1}{2v}=\dfrac{u+v}{2uv}$.

Step 2 — since $u+v=2y$, this gives $2y_1=\dfrac{2y}{2uv}=\dfrac{y}{uv}$, so $uv=\dfrac{y}{2y_1}$.

Step 3 — but $uv=\sqrt{(x+1)(x-1)}=\sqrt{x^2-1}$, so $\sqrt{x^2-1}=\dfrac{y}{2y_1}$, i.e. $2y_1\sqrt{x^2-1}=y$.

Step 4 — square both sides to remove the radical: $4y_1^2(x^2-1)=y^2$. …

Use the substitution $u=\sqrt{x+1},v=\sqrt{x-1}$ to reach $2y_1\sqrt{x^2-1}=y$, sq …

Trying to differentiate $2y=\sqrt{x+1}+\sqrt{x-1}$ twice directly without the $uv$-simplification trick — it tur …