SATHEE: Differential Equations 1 Question 16

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.

(d) Statement I is false; Statement II is true.

16.

Let a solution $y = y (x)$ of the differential equation

$x \sqrt{x^{2} - 1} d y - y \sqrt{y^{2} - 1} d x = 0 satisfy y (2) = \frac{2}{\sqrt{3}}$

Statement I $y (x) = \sec \sec^{- 1} x - \frac{π}{6}$ 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, $\frac{d y}{d x} = \frac{y \sqrt{y^{2} - 1}}{x \sqrt{x^{2} - 1}}$

$\int \frac{d y}{y \sqrt{y^{2} - 1}} = \int \frac{d x}{x \sqrt{x^{2} - 1}}$

$\Rightarrow \sec^{- 1} y = \sec^{- 1} x + c$

$\begin{aligned} At x = 2, y = \frac{2}{\sqrt{3}}; \frac{π}{6} = \frac{π}{3} + c \\ \Rightarrow c = - \frac{π}{6} \end{aligned}$

Now, $y = \sec (\sec^{- 1} x - \frac{π}{6})$

$\begin{aligned} = \cos [\cos^{- 1} \frac{1}{x} - \cos^{- 1} \frac{\sqrt{3}}{2}] \\ = \cos [\cos^{- 1} (\frac{\sqrt{3}}{2 x} + \sqrt{1 - \frac{1}{x^{2}}} \sqrt{1 - \frac{3}{4}})] \\ y & = \frac{\sqrt{3}}{2 x} + \frac{1}{2} \sqrt{1 - \frac{1}{x^{2}}} \end{aligned}$

Differential Equations 1 Question 16

16.

Answer:

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Differential Equations 1 Question 16

16.

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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