For the function \( y(x)=\sqrt{x+1}-\sqrt{x-1} \), which of the following are true?
(A) \( \left(\displaystyle\frac{dy}{dx}\right)_{at\,x=2}=\displaystyle\frac{1-\sqrt{3}}{2\sqrt{3}} \)
(B) \( (x^2-1)\displaystyle\frac{d^2y}{dx^2}-\displaystyle\frac{dy}{dx}+4y=0 \)
(C) \( (x^2-1)\left(\displaystyle\frac{dy}{dx}\right)^2-4y^2=0 \)
(D) If \( P(x)\displaystyle\frac{d^2y}{dx^2}+Q(x)\displaystyle\frac{dy}{dx}+\alpha y=0 \), then \( P(x)-xQ(x)-4\alpha=0 \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A) and (D) only
The derivative can be written in terms of y itself:
\( \displaystyle \frac{dy}{dx}=\frac{1}{2\sqrt{x+1}}-\frac{1}{2\sqrt{x-1}} \)
\( \displaystyle =\frac{\sqrt{x-1}-\sqrt{x+1}}{2\sqrt{x^2-1}} \)
\( \displaystyle =-\frac{y}{2\sqrt{x^2-1}} \)
So \( 2\sqrt{x^2-1}\,\frac{dy}{dx}=-y \). Squaring:
\( \displaystyle 4(x^2-1)\left(\frac{dy}{dx}\right)^2=y^2 \)
Differentiating this again with respect to x:
\( \displaystyle 8x\left(\frac{dy}{dx}\right)^2+8(x^2-1)\frac{dy}{dx}\frac{d^2y}{dx^2}=2y\frac{dy}{dx} \)
Dividing by \( 2\frac{dy}{dx} \) (which is never 0 here):
\( \displaystyle 4(x^2-1)\frac{d^2y}{dx^2}+4x\frac{dy}{dx}-y=0 \)
At x = 2:
\( \displaystyle \left(\frac{dy}{dx}\right)_{x=2}=\frac{1}{2\sqrt{3}}-\frac{1}{2} \)
\( \displaystyle =\frac{1-\sqrt{3}}{2\sqrt{3}} \)
So \( \left(\frac{dy}{dx}\right)_{x=2}=\frac{1-\sqrt{3}}{2\sqrt{3}} \), \( (x^2-1)\left(\frac{dy}{dx}\right)^2=\frac{y^2}{4} \) and the second-order equation is \( 4(x^2-1)\frac{d^2y}{dx^2}+4x\frac{dy}{dx}-y=0 \).
Checking the statements
(A) \( \left(\frac{dy}{dx}\right)_{at\,x=2}=\frac{1-\sqrt{3}}{2\sqrt{3}} \) — Correct. \( \frac{1}{2\sqrt{3}}-\frac{1}{2}=\frac{1-\sqrt{3}}{2\sqrt{3}} \).
(B) \( (x^2-1)\frac{d^2y}{dx^2}-\frac{dy}{dx}+4y=0 \) — Incorrect. Subtracting the true equation \( (x^2-1)\frac{d^2y}{dx^2}+x\frac{dy}{dx}-\frac{y}{4}=0 \) leaves \( -(x+1)\frac{dy}{dx}+\frac{17}{4}y \), which is positive since y > 0 and \( \frac{dy}{dx}\lt 0 \).
(C) \( (x^2-1)\left(\frac{dy}{dx}\right)^2-4y^2=0 \) — Incorrect. \( (x^2-1)\left(\frac{dy}{dx}\right)^2=\frac{y^2}{4} \), not \( 4y^2 \).
(D) If \( P(x)\frac{d^2y}{dx^2}+Q(x)\frac{dy}{dx}+\alpha y=0 \), then \( P(x)-xQ(x)-4\alpha=0 \) — Correct. The equation is k times the one above, so \( P=4k(x^2-1) \), \( Q=4kx \), \( \alpha=-k \), and \( 4k(x^2-1)-4kx^2+4k=0 \).