SATHEE: Differential Equations 2 Question 22

Differential Equations 2 Question 22

23.

If $y (x)$ satisfies the differential equation $y^{'} - y \tan x = 2 x \sec x$ and $y (0)$ , then

(2012)

(a) $y (\frac{π}{4}) = \frac{π^{2}}{8 \sqrt{2}}$

(b) $y^{'} (\frac{π}{4}) = \frac{π^{2}}{18}$

(d) $y^{'} (\frac{π}{3}) = \frac{4 π}{3} + \frac{2 π^{2}}{3 \sqrt{3}}$

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Answer:

Correct Answer: 23. (a, d)

Solution:

PLAN: Linear differential equation under one variable.

$\frac{d y}{d x} + P y = Q; I F = e^{\int P d x}$

$∴ Solution is, y (I F) = \int Q \cdot$ (IF) d x+C $

$y^{'} - y \tan x = 2 x \sec x and y (0) = 0$

$∴ \frac{d y}{d x} - y \tan x = 2 x \sec x$

$∴ I F = \int e^{- \tan x} d x = e^{\log | \cos x |} = \cos x$

Solution is $y \cdot \cos x = \int 2 x \sec x \cdot \cos x d x + C$

$\Rightarrow y \cdot \cos x = x^{2} + C$

$As y (0) = 0 \Rightarrow C = 0$

$∴ y = x^{2} \sec x$

$Now, y (\frac{π}{4}) = \frac{π^{2}}{8} \sqrt{2}$

$\Rightarrow y^{'} (\frac{π}{4}) = \frac{π}{\sqrt{2}} + \frac{π^{2}}{8 \sqrt{2}}$

$y (\frac{π}{3}) = \frac{2 π^{2}}{9} \Rightarrow y^{'} (\frac{π}{3}) = \frac{4 π}{3} + \frac{2 π^{2}}{3 \sqrt{3}}$

Differential Equations 2 Question 22

23.

Answer:

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Differential Equations 2 Question 22

23.

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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