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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Differential Equations. Please show all work.
![**Solving Initial Value Problems (I.V.P.) by Substitution**
To solve the given initial value problem using the substitution \( x = vy \):
\[ y \, dx + \left( y \cos\left(\frac{x}{y}\right) - x \right) dy = 0, \quad y(0) = 2 \]
we will use the following substitution:
\[ x = vy \]
Using this substitution, we will convert the given differential equation into a simpler form that can be solved more easily.
We begin by differentiating \( x = vy \) with respect to \( y \):
\[ \frac{dx}{dy} = v + y \frac{dv}{dy} \]
Next, we substitute \( x = vy \) and \( \frac{dx}{dy} = v + y \frac{dv}{dy} \) into the original differential equation.
This substitution will allow us to rewrite the differential equation in terms of \( v \) and \( y \), potentially simplifying the solution process.
By carefully following the steps and making necessary algebraic manipulations, we can find the solution to the initial value problem.
In this particular case, you would proceed to solve the substituted equation and apply the initial condition \( y(0) = 2 \) to find the specific solution.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc635a993-7586-4ec3-b52d-6ef96e1ee12a%2F41b8acc2-216f-49c9-87d2-99e7438acdb6%2F0ycex4_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Solving Initial Value Problems (I.V.P.) by Substitution**
To solve the given initial value problem using the substitution \( x = vy \):
\[ y \, dx + \left( y \cos\left(\frac{x}{y}\right) - x \right) dy = 0, \quad y(0) = 2 \]
we will use the following substitution:
\[ x = vy \]
Using this substitution, we will convert the given differential equation into a simpler form that can be solved more easily.
We begin by differentiating \( x = vy \) with respect to \( y \):
\[ \frac{dx}{dy} = v + y \frac{dv}{dy} \]
Next, we substitute \( x = vy \) and \( \frac{dx}{dy} = v + y \frac{dv}{dy} \) into the original differential equation.
This substitution will allow us to rewrite the differential equation in terms of \( v \) and \( y \), potentially simplifying the solution process.
By carefully following the steps and making necessary algebraic manipulations, we can find the solution to the initial value problem.
In this particular case, you would proceed to solve the substituted equation and apply the initial condition \( y(0) = 2 \) to find the specific solution.
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