(ex+y)dx+(2+x+ye³)dy = 0
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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please show steop by step explaining prob
![## Differential Equation Example
### Problem:
Consider the following first-order differential equation:
\[
(e^x + y)dx + (2 + x + ye^y)dy = 0
\]
### Explanation:
This is a partial differential equation where the terms involve both \( dx \) and \( dy \). The equation can be expressed as \( M(x, y)dx + N(x, y)dy = 0 \), where:
- \( M(x, y) = e^x + y \)
- \( N(x, y) = 2 + x + ye^y \)
### Goal:
Solve this differential equation for a possible explicit solution form.
### Method:
1. **Identify the type of differential equation**: In this case, we check if it is exact by verifying if the partial derivatives satisfy the condition \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \).
2. **Solve the exact equation**: If the equation is exact, find a potential function \( \psi(x, y) \) such that
\[
\frac{\partial \psi}{\partial x} = M(x, y)
\]
and
\[
\frac{\partial \psi}{\partial y} = N(x, y)
\]
3. **Integrate and find the solution**: Integrate both expressions with respect to their respective variables to find \( \psi(x, y) \).
### Learning Outcomes:
- Understanding and solving exact differential equations.
- Integrating partial derivatives to find a potential function.
- Applying criteria for exactness in differential equations.
This equation serves as a practical example illustrating how to approach and solve a differential equation of this type.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6aad8a00-c703-485c-bf76-99694ae31b4b%2F8fe41ecc-8d9a-4b1a-9a71-3a391db39e95%2F1sb7pkm_processed.png&w=3840&q=75)
Transcribed Image Text:## Differential Equation Example
### Problem:
Consider the following first-order differential equation:
\[
(e^x + y)dx + (2 + x + ye^y)dy = 0
\]
### Explanation:
This is a partial differential equation where the terms involve both \( dx \) and \( dy \). The equation can be expressed as \( M(x, y)dx + N(x, y)dy = 0 \), where:
- \( M(x, y) = e^x + y \)
- \( N(x, y) = 2 + x + ye^y \)
### Goal:
Solve this differential equation for a possible explicit solution form.
### Method:
1. **Identify the type of differential equation**: In this case, we check if it is exact by verifying if the partial derivatives satisfy the condition \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \).
2. **Solve the exact equation**: If the equation is exact, find a potential function \( \psi(x, y) \) such that
\[
\frac{\partial \psi}{\partial x} = M(x, y)
\]
and
\[
\frac{\partial \psi}{\partial y} = N(x, y)
\]
3. **Integrate and find the solution**: Integrate both expressions with respect to their respective variables to find \( \psi(x, y) \).
### Learning Outcomes:
- Understanding and solving exact differential equations.
- Integrating partial derivatives to find a potential function.
- Applying criteria for exactness in differential equations.
This equation serves as a practical example illustrating how to approach and solve a differential equation of this type.
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