Exact Equations

Exact equations are differential equations of the following form

They can also be written in the following forms:

Checking for exact equations

To check if an equation is exact, you have to take the partial derivatives of $M(x, y)$ and $N(x, y)$ and ensure that they are equal.

Solving Exact Equations

The solution to a exact equation is a function $I(x, y)$ whose partial derivatives can be put in place of $M(x,y)$ and $N(x, y)$

To find this function $I$, you can either integrate $\int Mdx$ or $\int Ndy$.

If $I(x, y)$ has partial derivatives $\frac{\partial I}{\partial x}$ and $\frac{\partial I}{\partial y}$, then $I(x, y) = C$ is a solution to the exact equation

$$$$

\frac{\partial I}{\partial x}dx + \frac{\partial I}{\partial y} dy=0

Examples

Example 1: Solve $(3x^2y^3 - 5x^4)dx + (y+3x^3y^2)dy$

Solution 1: This looks like an exact equation, but check to make sure.

• $M(x, y) = 3x^2y^3 - 5x^4$
• $N(x, y) = y + 3x^3y^2$
• $\frac{\partial M}{\partial y}=9x^2y^2$
• $\frac{\partial N}{\partial x}=9x^2y^2$

The partial derivatives are equal, so this is an exact equation.

We will integrate $\int M dx$ to find $I(x, y)$.

To find $f(y)$, remember that $\frac{\partial I}{\partial y} =N(x, y)$.

That means that $3x^3y^2 + f'(y)=y+ 3x^3y^2$. Simplify that to $f'(y) = y$, and you get $f(y) = \int f'(y) = \frac{y^2}{2} + C$.

$I(x, y) = x^3y^3 - x^5 + \frac{y^2}{2} + C$.

The solution then is $x^3y^3 - x^5 + \frac{y^2}{2}=C$. You can solve this for $y$ to get the explicit solution.