termine if the differential equation is exact, and fin
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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![**Title:** Exact Differential Equations and Their Solutions
**Content:**
**Problem Statement:**
Determine if the differential equation is exact, and find the solution.
\[ (\cos(x) + e^{y+x} )dx + \left( 1 + e^{y+x} + 2y e^{y^2} \right) dy = 0 \]
**Analysis:**
To determine if the given differential equation is exact, we need to check if there exists a function \( F(x, y) \) such that:
\[ \frac{\partial F}{\partial x} = \cos(x) + e^{y+x} \]
\[ \frac{\partial F}{\partial y} = 1 + e^{y+x} + 2y e^{y^2} \]
For the differential equation to be exact, the mixed partial derivatives of \( F \) must be equal:
\[ \frac{\partial}{\partial y} \left( \frac{\partial F}{\partial x} \right) = \frac{\partial}{\partial x} \left( \frac{\partial F}{\partial y} \right) \]
First, calculate \( \frac{\partial}{\partial y} (\cos(x) + e^{y+x}) \):
\[ \frac{\partial}{\partial y} (\cos(x) + e^{y+x}) = 0 + e^{y+x} = e^{y+x} \]
Next, calculate \( \frac{\partial}{\partial x} (1 + e^{y+x} + 2y e^{y^2}) \):
\[ \frac{\partial}{\partial x} (1 + e^{y+x} + 2y e^{y^2}) = e^{y+x} + 0 = e^{y+x} \]
Since both partial derivatives are equal, the differential equation is exact.
**Solution:**
To find the solution, integrate \( \cos(x) + e^{y+x} \) with respect to \( x \):
\[ F(x,y) = \int (\cos(x) + e^{y+x}) \, dx = \sin(x) + e^{y+x} + g(y) \]
Then, differentiate \( F(x,y) \) with respect to \( y \) and set the expression equal to \( 1 + e^{y+x} + 2y e](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8bdf6582-40e2-47b9-bd75-44351bba3f76%2F8bf0b6c5-86a4-44ab-966a-1736b090feab%2Fs0fmofx_processed.png&w=3840&q=75)
Transcribed Image Text:**Title:** Exact Differential Equations and Their Solutions
**Content:**
**Problem Statement:**
Determine if the differential equation is exact, and find the solution.
\[ (\cos(x) + e^{y+x} )dx + \left( 1 + e^{y+x} + 2y e^{y^2} \right) dy = 0 \]
**Analysis:**
To determine if the given differential equation is exact, we need to check if there exists a function \( F(x, y) \) such that:
\[ \frac{\partial F}{\partial x} = \cos(x) + e^{y+x} \]
\[ \frac{\partial F}{\partial y} = 1 + e^{y+x} + 2y e^{y^2} \]
For the differential equation to be exact, the mixed partial derivatives of \( F \) must be equal:
\[ \frac{\partial}{\partial y} \left( \frac{\partial F}{\partial x} \right) = \frac{\partial}{\partial x} \left( \frac{\partial F}{\partial y} \right) \]
First, calculate \( \frac{\partial}{\partial y} (\cos(x) + e^{y+x}) \):
\[ \frac{\partial}{\partial y} (\cos(x) + e^{y+x}) = 0 + e^{y+x} = e^{y+x} \]
Next, calculate \( \frac{\partial}{\partial x} (1 + e^{y+x} + 2y e^{y^2}) \):
\[ \frac{\partial}{\partial x} (1 + e^{y+x} + 2y e^{y^2}) = e^{y+x} + 0 = e^{y+x} \]
Since both partial derivatives are equal, the differential equation is exact.
**Solution:**
To find the solution, integrate \( \cos(x) + e^{y+x} \) with respect to \( x \):
\[ F(x,y) = \int (\cos(x) + e^{y+x}) \, dx = \sin(x) + e^{y+x} + g(y) \]
Then, differentiate \( F(x,y) \) with respect to \( y \) and set the expression equal to \( 1 + e^{y+x} + 2y e
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