User:Egm6321.f09.Team1.vasquez/Homework3

= Problem 4 - Exactness condition for $$\displaystyle n=1$$ =

Given
A first order differential equation


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$$
 * $$F(x,y,y')=0=\frac{d \phi}{dx}(x,y)$$
 * $$\displaystyle (Eq. 1)
 * $$\displaystyle (Eq. 1)
 * }
 * }

Find
4.1) Find $$\displaystyle f_0$$ in terms of $$\displaystyle \phi$$

4.2) Find $$\displaystyle f_1$$ in terms of $$\displaystyle \phi$$, i.e. show that $$ \displaystyle f_1=\phi_y$$

4.3) Show that $$ f_0-\frac{df_1}{dx}=0 \Leftrightarrow \phi_{xy}=\phi_{yx}$$

Solution
The exactness condition for a $$n^{th}$$-order differential equation is given by


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f_0 - \frac + \frac -  \cdots  + ( - 1)^n \frac = 0 $$ $$
 * $$ \displaystyle
 * $$ \displaystyle
 * $$\displaystyle (Eq. 2)
 * }
 * }

where


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f_i = \frac \quad \quad i = 1, \ldots ,n$$ $$
 * $$ \displaystyle
 * $$ \displaystyle
 * $$\displaystyle (Eq. 3)
 * }
 * }

Expanding equation (1) gives


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F=\frac{d \phi}{dx} = \frac{\partial \phi}{\partial x} + \frac{\partial \phi}{\partial y}\frac{\partial y}{\partial x} = \phi_x + \phi_y y' $$ $$
 * $$ \displaystyle
 * $$ \displaystyle
 * $$\displaystyle (Eq. 4)
 * }
 * }

Since $$\displaystyle n=1$$, we only need to compute $$\displaystyle f_0, f_1$$ as follows:


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f_0 = \frac = \phi _{xy}  + \phi _{yy} y' + \phi_y \cancelto{0}{\left( \frac \right)} = \phi _{xy}  + \phi _{yy} y' $$ $$
 * $$ \displaystyle
 * $$ \displaystyle
 * $$\displaystyle (Eq. 5)
 * }
 * }


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f_1 = \frac = \phi _{y} $$ $$
 * $$ \displaystyle
 * $$ \displaystyle
 * $$\displaystyle (Eq. 6)
 * }
 * }

Now, we can take the total derivative of $$\displaystyle f_1 $$


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\frac{d f_1}{dx} = \frac{\partial f_1}{\partial x} + \frac{\partial f_1}{\partial y}\frac{\partial y}{\partial x} $$ $$
 * $$ \displaystyle
 * $$ \displaystyle
 * $$\displaystyle (Eq. 7)
 * }
 * }

Substituting (6) into (7) yields


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\frac{d f_1}{dx} = \phi_{yx} + \phi_{yy} y' $$ $$
 * $$ \displaystyle
 * $$ \displaystyle
 * $$\displaystyle (Eq. 8)
 * }
 * }

Substituting (5) and (8) into the exactness condition gives


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\left(\phi _{xy} + \cancel{\phi _{yy} y'}\right) - \left(\phi_{yx} + \cancel{\phi_{yy} y'}\right)=0 $$ $$
 * $$ \displaystyle
 * $$ \displaystyle
 * $$\displaystyle (Eq. 8)
 * }
 * }

Then the exactness condition is satisfied if


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\phi _{xy} = \phi_{yx} $$ $$
 * $$ \displaystyle
 * $$ \displaystyle
 * $$\displaystyle (Eq. 9)
 * }
 * }