Differential Equations 1 Question 16
16. Let a solution $y=y(x)$ of the differential equation
$$ x \sqrt{x^{2}-1} d y-y \sqrt{y^{2}-1} d x=0 \text { satisfy } y(2)=\frac{2}{\sqrt{3}} $$
Statement I $y(x)=\sec \sec ^{-1} x-\frac{\pi}{6}$ and
Statement II $y(x)$ is given by $\frac{1}{y}=\frac{2 \sqrt{3}}{x}-\sqrt{1-\frac{1}{x^{2}}}$
$(2008,3 M)$
Analytical & Descriptive Questions
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Answer:
Correct Answer: 16. (c)
Solution:
- Given, $\quad \frac{d y}{d x}=\frac{y \sqrt{y^{2}-1}}{x \sqrt{x^{2}-1}}$
$$ \begin{aligned} \quad \int \frac{d y}{y \sqrt{y^{2}-1}} & =\int \frac{d x}{x \sqrt{x^{2}-1}} \\ \Rightarrow \quad \sec ^{-1} y & =\sec ^{-1} x+c \end{aligned} $$
$$ \begin{aligned} & \text { At } x=2, y=\frac{2}{\sqrt{3}} ; \frac{\pi}{6}=\frac{\pi}{3}+c \\ & \Rightarrow c=-\frac{\pi}{6} \end{aligned} $$
Now, $y=\sec \sec ^{-1} x-\frac{\pi}{6}$
$$ \begin{aligned} & =\cos \cos ^{-1} \frac{1}{x}-\cos ^{-1} \frac{\sqrt{3}}{2} \\ & =\cos \cos ^{-1} \frac{\sqrt{3}}{2 x}+\sqrt{1-\frac{1}{x^{2}}} \sqrt{1-\frac{3}{4}} \\ y & =\frac{\sqrt{3}}{2 x}+\frac{1}{2} \sqrt{1-\frac{1}{x^{2}}} \end{aligned} $$
