7  L7: Exact Equations (2.5)

7.1 Definition: pre-Exact Equations

Definition:

A potential function \(F(x,y)\) is a function of two independent variables.

For a fixed constant \(C\), the level curves (equipotential curves) \[F(x,y)=C\] represent the points in the \(x,y\) plane with the same potential \(C\).

Definition:

The total differential of \(F(x,y)\) is¹

\[ dF =\frac{\partial F}{\partial x}\,dx+\frac{\partial F}{\partial y}\,dy \]

7.1.1 Level curves

7.1.2 Example (Level Newbie)

Find the total differential of

\[F(x,y)= x^2+xy+y^2\]

Solution:

\[dF =(2x+y)\,dx+(x+2y)\,dy\]

7.1.3 Note:

What is total differential for a level curve? Can I calculate \(dy/dx\) from an implicit

OJO

If we are on a level curve: \[\Rightarrow F(x,y)=C\]

\[\begin{align} \Rightarrow \\ dF &=\frac{\partial F}{\partial x}\,dx+\frac{\partial F}{\partial y}\,dy=0\\ \Rightarrow \\\\ \frac{dy}{dx}&= -\frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial y}}\\ &=g(x,y) \end{align}\]

provided \(\frac{\partial F}{\partial y}\neq 0\)

7.2 Definition: exact diffQ

Definition:

If a function \(F(x,y)\) exist such that \[ \frac{\partial F}{\partial x}=M(x,y)\quad\text{and}\quad\frac{\partial F}{\partial y}=N(x,y)\] then \[M(x,y)\,dx+N(x,y)\,dy=0\] is said to be an exact diffyQ²

7.3 Test for exactness

Test:

\[M(x,y)\,dx+N(x,y)\,dy=0\] is exact if and only if³

\[\boxed{\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}}\]

7.4 Method for non-separable, non-linear:

For solving diffyQ of the form

\[\frac{dy}{dx}=g(x,y)\] If \[dF=0\Rightarrow\]

\[ \frac{dy}{dx}=-\frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial y}} = -\frac{M(x,y)}{N(x,y)}\]

Method:

1. Write in differential form so that

\[\boxed{M(x,y)\,dx+N(x,y)\,dy=0}\]

2. Make sure the equation is exact⁴

\[\boxed{\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}}\]

3. Solve for \(F(x,y)\)

\[\frac{\partial F}{\partial x}=M(x,y)\] \[\Rightarrow \boxed{F(x,y)=\int M(x,y)\,dx+\phi(y)}\]

4. Solve for the unknown function \(\phi(y)\)

We know that: \[\frac{\partial F}{\partial y}=N(x,y)\] from 3) we know how much if \(F\), then

\[N=\frac{\partial}{\partial y}\left(\int M(x,y)\,dx+\phi(y)\right)\] \[\phi'(y)=N-\frac{\partial}{\partial y}\left(\int M(x,y)\,dx\right)\]

Therefore:

\[\boxed{\phi(y)=\int \left[N-\frac{\partial}{\partial y}\left(\int M\,dx\right)\right]\,dy}\]

5. Write implicit general solution:

\[\boxed{F(x,y)=C}\]

7.5 Example step by step

Solve: \[\frac{dy}{dx}=\frac{e^y+x}{e^{2y}-xe^y} \]

Solution

Sol:⁵

\[M(x,y)=F_x =e^y+x\] \[N(x,y)=-F_y =-(e^{2y}-xe^y)=xe^y-e^{2y}\]

1. \[(e^y+x)\,dx+(xe^y-e^{2y})\,dy=0\]

2. Exact? \[M_y=N_x\,?\] \[M_y=e^y,\quad N_x=e^y\] Then \[\checkmark\]

3. \[\frac{\partial F}{\partial x}=M=e^y+x\] \[F=\int (e^y+x)\,dx +\phi(y)\] \[F=xe^y+\frac{x^2}{2} +\phi(y)\]

4. \[\frac{\partial F}{\partial y}=N=xe^y-e^{2y}\] Then

\[xe^y-e^{2y}= \frac{\partial }{\partial y}\left(xe^y+\frac{x^2}{2} +\phi(y)\right)\]

\[xe^y-e^{2y}= xe^y+0+\phi'(y)\] \[\phi'(y)=-e^{2y}\] \[\phi(y)=-\frac{e^{2y}}{2}\]

5. Implicit solution:

\[F(x,y)=xe^y+\frac{x^2}{2}-\frac{e^{2y}}{2}=C\]

7.6 References

1. Khan: Exact diffyQ Example 1 https://youtu.be/Pb04ntcDJcQ
2. Khan: Exact diffyQ Example 2 https://youtu.be/utQi1ZhF__Q
3. Khan: Exact diffyQ Example 3 https://youtu.be/eu_GFuU7tLI
4. Wiki: Exact diffyQ https://en.wikipedia.org/wiki/Exact_differential_equation
5. Wiki: Second partial derivatives https://www.khanacademy.org/math/multivariable-calculus/multivariable-derivatives/partial-derivative-and-gradient-articles/a/second-partial-derivatives

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1. true for well behave functions, \(\Delta z \approx dz = F_x(x,y)dx+F_y(x,y)dy\)

2. if we are on a level curve should not exist a change on height.(\(F=C \rightarrow dF=0\))

3. If \(F_{xy}\) and \(F_{yx}\) are continuous then the mix partial derivatives are equal(\(F_{xy}=F_{yx}\))

4. \(F_{yx}=F_{xy}\) ?

5. \(F_x\equiv \frac{\partial F}{\partial x}\)