# Exact Equations

Author: John J Weber III, PhD Corresponding Textbook Sections:

• Section 2.4 – Exact Equations

## Expected Educational Results

• Objective 6–1: I can identify the exact differential form.

• Objective 6–2: I can determine if an ODE is an exact equation.

• Objective 6–3: I can solve exact equations.

## Exact Equations

### Method of Solving Exact Equations

1. Identify $M\left(x,y\right)$$M(x,y)$ from the exact differential form.

2. Identify $N\left(x,y\right)$$N(x,y)$ from the exact differential form.

3. Find $\frac{\partial M\left(x,y\right)}{\partial y}$$\displaystyle \frac{\partial M(x,y)}{\partial y}$.

4. Find $\frac{\partial N\left(x,y\right)}{\partial x}$$\displaystyle \frac{\partial N(x,y)}{\partial x}$.

5. Compare $\frac{\partial M\left(x,y\right)}{\partial y}$$\displaystyle \frac{\partial M(x,y)}{\partial y}$ and $\frac{\partial N\left(x,y\right)}{\partial x}$$\displaystyle \frac{\partial N(x,y)}{\partial x}$.

• If $\frac{\partial M\left(x,y\right)}{\partial y}=\frac{\partial N\left(x,y\right)}{\partial x}$$\displaystyle \frac{\partial M(x,y)}{\partial y}=\frac{\partial N(x,y)}{\partial x}$, then the differential equation is an exact equation, proceed to next step.

• If $\frac{\partial M\left(x,y\right)}{\partial y}\ne \frac{\partial N\left(x,y\right)}{\partial x}$$\displaystyle \frac{\partial M(x,y)}{\partial y}\ne\frac{\partial N(x,y)}{\partial x}$, then we need another method to be discussed in a future class period.

6. Since we know partial derivatives of $F\left(x,y\right)$$F(x,y)$, we need to integrate the partial derivatives to find $F\left(x,y\right)$$F(x,y)$:

• Since $M\left(x,y\right)=\frac{\partial F\left(x,y\right)}{\partial x}$$\displaystyle M(x,y)=\frac{\partial F(x,y)}{\partial x}$, to find $F\left(x,y\right)$$F(x,y)$ evaluate: $\int M\left(x,y\right)\phantom{\rule{0.167em}{0ex}}dx$$\displaystyle \int{M(x,y)}\,dx$, OR

• Since $N\left(x,y\right)=\frac{\partial F\left(x,y\right)}{\partial y}$$\displaystyle N(x,y)=\frac{\partial F(x,y)}{\partial y}$, to find $F\left(x,y\right)$$F(x,y)$ evaluate: $\int N\left(x,y\right)\phantom{\rule{0.167em}{0ex}}dy$$\displaystyle \int{N(x,y)}\,dy$

7. Since we are evaluating indefinite integrals, the antiderivatives require a constant term:

• When integrating with respect to $x$$x$, $y$$y$ is a constant; evaluate $F\left(x,y\right)=\int M\left(x,y\right)\phantom{\rule{0.167em}{0ex}}dx+g\left(y\right)$$\displaystyle F(x,y)=\int{M(x,y)}\,dx+g(y)$; OR

• When integrating with respect to $y$$y$, $x$$x$ is a constant; evaluate $F\left(x,y\right)=\int N\left(x,y\right)\phantom{\rule{0.167em}{0ex}}dy+g\left(x\right)$$\displaystyle F(x,y)=\int{N(x,y)}\,dy+g(x)$

8. Find the constant of integration:

1. Case 1

• Evaluate: $\frac{\partial F\left(x,y\right)}{\partial y}=\frac{\partial \left[\int M\left(x,y\right)\phantom{\rule{0.167em}{0ex}}dx+g\left(y\right)\right]}{\partial y}=M\left(x,y\right)+{g}^{\phantom{\rule{0.167em}{0ex}}\prime }\left(y\right)$$\displaystyle \frac{\partial F(x,y)}{\partial y}=\frac{\partial \left[\int{M(x,y)}\,dx+g(y)\right]}{\partial y}=M(x,y)+g^{\,\prime}(y)$ [which is equivalent to $N\left(x,y\right)$$N(x,y)$]

• Solve $M\left(x,y\right)+{g}^{\phantom{\rule{0.167em}{0ex}}\prime }\left(y\right)=N\left(x,y\right)$$M(x,y)+g^{\,\prime}(y)=N(x,y)$ for ${g}^{\phantom{\rule{0.167em}{0ex}}\prime }\left(y\right)$$g^{\,\prime}(y)$

• $g\left(y\right)=\int {g}^{\phantom{\rule{0.167em}{0ex}}\prime }\left(y\right)\phantom{\rule{0.167em}{0ex}}dy$$\displaystyle g(y) = \int{g^{\,\prime}(y)\,dy}$

2. Case 2

• Evaluate: $\frac{\partial F\left(x,y\right)}{\partial x}=\frac{\partial \left[\int N\left(x,y\right)\phantom{\rule{0.167em}{0ex}}dy+g\left(x\right)\right]}{\partial x}=N\left(x,y\right)+{g}^{\phantom{\rule{0.167em}{0ex}}\prime }\left(x\right)$$\displaystyle \frac{\partial F(x,y)}{\partial x}=\frac{\partial \left[\int{N(x,y)}\,dy+g(x)\right]}{\partial x}=N(x,y)+g^{\,\prime}(x)$ [which is equivalent to $M\left(x,y\right)$$M(x,y)$]

• Solve $N\left(x,y\right)+{g}^{\phantom{\rule{0.167em}{0ex}}\prime }\left(x\right)=M\left(x,y\right)$$N(x,y)+g^{\,\prime}(x)=M(x,y)$ for ${g}^{\phantom{\rule{0.167em}{0ex}}\prime }\left(x\right)$$g^{\,\prime}(x)$

• $g\left(x\right)=\int {g}^{\phantom{\rule{0.167em}{0ex}}\prime }\left(x\right)\phantom{\rule{0.167em}{0ex}}dx$$\displaystyle g(x) = \int{g^{\,\prime}(x)\,dx}$

9. Thus, $F\left(x,y\right)=\int M\left(x,y\right)\phantom{\rule{0.167em}{0ex}}dx+g\left(y\right)$$\displaystyle F(x,y)=\int{M(x,y)}\,dx+g(y)$ OR $F\left(x,y\right)=\int N\left(x,y\right)\phantom{\rule{0.167em}{0ex}}dy+g\left(x\right)$$\displaystyle F(x,y)=\int{N(x,y)}\,dy+g(x)$ usually, written as $F\left(x,y\right)=C$$F(x,y)=C$