Differential Equations 1 Question 16
Assertion and Reason
For the following question, choose the correct answer from the codes (a), (b), (c) and (d) defined as follows.
(a) Statement I is true, Statement II is also true; Statement II is the correct explanation of Statement I.
(b) Statement I is true, Statement II is also true; Statement II is not the correct explanation of Statement I.
(c) Statement I is true; Statement II is false.
(d) Statement I is false; Statement II is true.
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 $\frac{1}{y}=\frac{2 \sqrt{3}}{x}-\sqrt{1-\frac{1}{x^{2}}}$
$(2008,3 M)$
Show Answer
Answer:
Correct Answer: 16. (b)
Solution:
- Given, $\quad \frac{d y}{d x}=\frac{y \sqrt{y^{2}-1}}{x \sqrt{x^{2}-1}}$
$\int \frac{d y}{y \sqrt{y^{2}-1}} =\int \frac{d x}{x \sqrt{x^{2}-1}} $
$\Rightarrow \quad \sec ^{-1} y =\sec ^{-1} x+c$
$ \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} $
