Solve the separable equation y(3+ sin(x²))dy = xcos(x?)dx subject to y(0) = -Vā where a > 0 is aconstant.Select one:y(x) = VIn(3 + sin(x-)) + a – In(3)= tyIn(3 + sin(xx²)) + a – In(3)(x) = -VIn(3 + sin(x²)) –- a + In(3)(x) = -VIn(3 + sin(x²)) + a – In(3)

Answered: Solve the separable equation y(3+…

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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Question

Solve the separable equation y(3+ sin(x²))dy = xcos(x?)dx subject to y(0) = -Vā where a > 0 is a
constant.
Select one:
y(x) = VIn(3 + sin(x-)) + a – In(3)
= tyIn(3 + sin(xx²)) + a – In(3)
(x) = -VIn(3 + sin(x²)) –
- a + In(3)
(x) = -VIn(3 + sin(x²)) + a – In(3)

Expert Solution

Step 1

Given that,

$y (3 + \sin (x^{2})) d y = x \cos (x^{2}) d x$

Step 2

Simplification:

Since we have,

$y (3 + \sin (x^{2})) d y = x \cos (x^{2}) d x$

So using variable separable form

$y d y = \frac{x \cos (x^{2})}{(3 + \sin (x^{2}))} d x$

Integrate both sides,

$\int y d y = \int \frac{x \cos (x^{2})}{(3 + \sin (x^{2}))} d x$ ...(1)

Substitute $3 + \sin (x^{2}) = u \Rightarrow 2 x \cos (x^{2}) d x = d u \Rightarrow x \cos (x^{2}) d x = \frac{d u}{2}$ in (1),

$\int y d y = \int \frac{1}{u} \frac{d u}{2} \int y d y = \frac{1}{2} \int \frac{1}{u} d u \frac{y^{2}}{2} = \frac{1}{2} \ln u + \ln k y^{2} = \ln u + \ln c y = \pm \sqrt{\ln u + \ln c} y = \pm \sqrt{\ln (3 + \sin (x^{2})) + \ln c} . . . (2)$

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Solve the separable equation y(3+ sin(x²))dy = xcos(x?)dx subject to y(0) = -Vā where a > 0 is a constant. Select one: y(x) = VIn(3 + sin(x-)) + a – In(3) = tyIn(3 + sin(xx²)) + a – In(3) (x) = -VIn(3 + sin(x²)) – - a + In(3) (x) = -VIn(3 + sin(x²)) + a – In(3)