SATHEE: Differential Equations 2 Question 3

Differential Equations 2 Question 3

3.

Let $y = y (x)$ be the solution of the differential equation, $\frac{d y}{d x} + y \tan x = 2 x + x^{2} \tan x, x \in (- \frac{π}{2}, \frac{π}{2})$ , such that $y (0) = 1$ . Then

(2019 Main, 10 April II)

(a) $y^{'} (\frac{π}{4}) - y^{'} (- \frac{π}{4}) = π - \sqrt{2}$

(b) $y^{'} (\frac{π}{4}) + y^{'} (- \frac{π}{4}) = - \sqrt{2}$

(d) $y (\frac{π}{4}) - y (- \frac{π}{4}) = \sqrt{2}$

Show Answer

Answer:

Correct Answer: 3. (a)

Solution:

Given differential equation is

$\frac{d y}{d x} + y \tan x = 2 x + x^{2} \tan x$ , which is linear differential equation in the form of $\frac{d y}{d x} + P y = Q$ .

Here, $P = \tan x$ and $Q = 2 x + x^{2} \tan x$

$∴ I F = e^{\int \tan x d x} = e^{\log_{e} (\sec x)} = \sec x$

Now, solution of linear differential equation is given as

$\begin{aligned} y \times I F = \int (Q \times I F) d x + C \\ ∴ y (\sec x) = \int (2 x + x^{2} \tan x) \sec x d x + C \\ = \int (2 x \sec x) d x + \int x^{2} \sec x \tan x d x + C \\ ∵ \int x^{2} \sec x \tan x d x = x^{2} \sec x - \int (2 x \sec x) d x \end{aligned}$

Therefore, solution is

$y \sec x = 2 \int x \sec x d x + x^{2} \sec x - 2 \int x \sec x d x + C$

$\Rightarrow y \sec x = x^{2} \sec x + C \dots$ (i)

$∵ y (0) = 1 \Rightarrow 1 (1) = 0 (1) + C \Rightarrow C = 1$

Now, $y = x^{2} + \cos x$

and $y^{'} = 2 x - \sin x$

According to options,

$y^{'} (\frac{π}{4}) - y^{'} (\frac{- π}{4}) = (2 \frac{π}{4} - \frac{1}{\sqrt{2}})$

$- (2 (- \frac{π}{4}) + \frac{1}{\sqrt{2}}) = π - \sqrt{2}$

and $y^{'} (\frac{π}{4}) + y^{'} (- \frac{π}{4}) = 2 \frac{π}{4} - \frac{1}{\sqrt{2}} + 2 - \frac{π}{4} + \frac{1}{\sqrt{2}} = 0$

and $y (\frac{π}{4}) + y (- \frac{π}{4}) = \frac{π^{2}}{16} + \frac{1}{\sqrt{2}} + \frac{π^{2}}{16} + \frac{1}{\sqrt{2}} = \frac{π^{2}}{4} + \sqrt{2}$

and $y (\frac{π}{4}) - y (- \frac{π}{4}) = \frac{π^{2}}{16} + \frac{1}{\sqrt{2}} - \frac{π^{2}}{16} - \frac{1}{\sqrt{2}} = 0$

Differential Equations 2 Question 3

3.

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Differential Equations 2 Question 3

3.

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Menu