dy 3 Use the method of separation of variables to find the solution to =y dx dy= dx Begin by separating the variables.
dy 3 Use the method of separation of variables to find the solution to =y dx dy= dx Begin by separating the variables.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![In applications, most initial value problems will have a unique solution. In fact, the existence of unique solutions is so important that there is a theorem about the existence and uniqueness of a solution. Consider the initial value problem \( \frac{dy}{dx} = f(x,y), \, y(x_0) = y_0 \). If \( f \) and \( \frac{\partial f}{\partial y} \) are continuous functions in some rectangle \( R = \{(x,y): a < x < b, \, c < y < d\} \) that contains the point \( (x_0, y_0) \), then the initial value problem has a unique solution \( \phi(x) \) in some interval \( x_0 - \delta < x < x_0 + \delta \), where \( \delta \) is a positive number. The method for separable equations can give a solution, but it may not give all the solutions. To illustrate this, consider the equation \( \frac{dy}{dx} = \frac{1}{y^3} \). Answer parts (a) through (d).
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(a) Use the method of separation of variables to find the solution to \( \frac{dy}{dx} = \frac{1}{y^3} \). Begin by separating the variables.
\[ \int y^3 \, dy = \int dx \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F799db758-6264-477e-875f-1cfecc426da7%2Fc7358196-2f07-48f6-93f9-a86dd7e17111%2Febehc0f_processed.png&w=3840&q=75)
Transcribed Image Text:In applications, most initial value problems will have a unique solution. In fact, the existence of unique solutions is so important that there is a theorem about the existence and uniqueness of a solution. Consider the initial value problem \( \frac{dy}{dx} = f(x,y), \, y(x_0) = y_0 \). If \( f \) and \( \frac{\partial f}{\partial y} \) are continuous functions in some rectangle \( R = \{(x,y): a < x < b, \, c < y < d\} \) that contains the point \( (x_0, y_0) \), then the initial value problem has a unique solution \( \phi(x) \) in some interval \( x_0 - \delta < x < x_0 + \delta \), where \( \delta \) is a positive number. The method for separable equations can give a solution, but it may not give all the solutions. To illustrate this, consider the equation \( \frac{dy}{dx} = \frac{1}{y^3} \). Answer parts (a) through (d).
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(a) Use the method of separation of variables to find the solution to \( \frac{dy}{dx} = \frac{1}{y^3} \). Begin by separating the variables.
\[ \int y^3 \, dy = \int dx \]
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