SATHEE: Differential Equations 2 Question 12

Differential Equations 2 Question 12

12.

If $\frac{d y}{d x} + \frac{3}{\cos^{2} x} y = \frac{1}{\cos^{2} x}, x \in (\frac{- π}{3}, \frac{π}{3})$ and $y \frac{π}{4} = (\frac{4}{3})$ , then $y - \frac{π}{4}$ equals

(2019 Main, 10 Jan I)

(a) $\frac{1}{3} + e^{6}$

(b) $- \frac{4}{3}$

(d) $\frac{1}{3}$

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Answer:

Correct Answer: 12. (a)

Solution:

Given, differential equation is $\frac{d y}{d x} + (\frac{3}{\cos^{2} x}) y = \frac{1}{\cos^{2} x}$ , which is a linear differential equation of the form

$\frac{d y}{d x} + P y = Q$ , where $P = \frac{3}{\cos^{2} x}$ and $Q = \frac{1}{\cos^{2} x}$ .

Now, Integrating factor

$I F = e^{\int \frac{3}{\cos^{2} x} d x} = e^{β \sec^{2} x d x} = e^{3 \tan x}$ and the solution of differential equation is given by

$\begin{aligned} y (I F) = \int (Q . (I F)) d x \\ \Rightarrow e^{3 \tan x} \cdot y = \int e^{3 \tan x} \sec^{2} x d x \dots \dots . (i) \\ Let I = \int e^{3 \tan x} \sec^{2} x d x \\ Put 3 \tan x = t \\ \Rightarrow 3 \sec^{2} x d x = d t \\ ∴ I = \int \frac{e^{t}}{3} d t = \frac{e^{t}}{3} + C = \frac{e^{3 \tan x}}{3} + C \end{aligned}$

From Eq. (i)

$e^{3 \tan x} \cdot y = \frac{e^{3 \tan x}}{3} + C$

It is given that when,

$x = \frac{π}{4}, y is \frac{4}{3}$

$\Rightarrow e^{3} \frac{4}{3} = \frac{e^{3}}{3} + C$

$\Rightarrow C = e^{3}$

Thus, $e^{3 \tan x} y = \frac{e^{3 \tan x}}{3} + e^{3}$

Now, when $x = - \frac{π}{4}, e^{- 3} y = \frac{e^{- 3}}{3} + e^{3}$

$\Rightarrow y = e^{6} + \frac{1}{3}$ $[∵ \tan - \frac{π}{4} = - 1]$

Differential Equations 2 Question 12

12.

Answer:

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Differential Equations 2 Question 12

12.

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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