JEE Advanced · Mathematics · 29. Differential Eqns
Let a solution \(y=y(x)\) of the differential equation \(x \sqrt{x^2-1} d y-y \sqrt{y^2-1} d x=0\) satisfy \(y(2)=\frac{2}{\sqrt{3}}\)
Statement \(1 y(x)=\sec \left(\sec ^{-1} x-\frac{\pi}{6}\right)\)
Statement \(2 y(x)\) is given by \(\frac{1}{y}=\frac{2 \sqrt{3}}{x}-\sqrt{1-\frac{1}{x^2}}\)
- A
Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1.
- B
Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1.
- C
Statement 1 is true, Statement 2 is false.
- D
Statement 1 is false, Statement 2 is true
Answer & Solution
Correct Answer
(C)
Statement 1 is true, Statement 2 is false.
Step-by-step Solution
Detailed explanation
\(\because \frac{d y}{d x}=\frac{y \sqrt{y^2-1}}{x \sqrt{x^2-1}} \Rightarrow \int \frac{d y}{y \sqrt{y^2-1}}=\int \frac{d x}{x \sqrt{x^2-1}}\) \(\Rightarrow \quad \sec ^{-1} y=\sec ^{-1} x+C\) At \(\quad x=2, y=\frac{2}{\sqrt{3}}\) \[ \frac{\pi}{6}=\frac{\pi}{3}+C \Rightarrow C=-\frac{\pi}{6} \]
Now, \(y=\sec \left(\sec ^{-1} x-\frac{\pi}{6}\right)\)
$$
\begin{aligned}
& =\cos \left[\cos ^{-1} \frac{1}{x}-\cos ^{-1} \frac{\sqrt{3}}{2}\right] \\
& =\cos \left[\cos ^{-1}\left(\frac{\sqrt{3}}{2 x}+\sqrt{1-\frac{1}{x^2}} \sqrt{1-\frac{3}{4}}\right)\right] \\
\Rightarrow \quad \frac{1}{y} & =\frac{\sqrt{3}}{2 x}+\frac{1}{2} \sqrt{1-\frac{1}{x^2}} \\
Now, \(y=\sec \left(\sec ^{-1} x-\frac{\pi}{6}\right)\)
$$
\begin{aligned}
& =\cos \left[\cos ^{-1} \frac{1}{x}-\cos ^{-1} \frac{\sqrt{3}}{2}\right] \\
& =\cos \left[\cos ^{-1}\left(\frac{\sqrt{3}}{2 x}+\sqrt{1-\frac{1}{x^2}} \sqrt{1-\frac{3}{4}}\right)\right] \\
\Rightarrow \quad \frac{1}{y} & =\frac{\sqrt{3}}{2 x}+\frac{1}{2} \sqrt{1-\frac{1}{x^2}} \\
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