termine if the differential equation is exact, and fin

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
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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
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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