Exact ordinary differential equations

We will say that the ODE $P (x, y) d x + Q (x, y) d y = 0$ where $P, Q$ are functions of $x$ and $y$ , it is exact if $P_{y} = Q_{x}$ , where $P_{y}$ indicates the partial derivative of $P$ with respect to $y$ and $Q_{x}$ , the partial derivative of $Q$ with respect to $x$ .

Example

An example of an exact ODE would be: $(x^{3} + y^{3}) d x + 3 x y^{2} d y = 0$ In effect, calling $P (x, y) = (x^{3} + y^{3}), Q (x, y) = 3 x y^{2}$ we have $P_{y} = 3 y^{2} = Q_{x}$ .

It is necessary to observe that not all the ODEs are exact, for example

Example

$- y^{2} \cdot d x + (x^{2} + x y) \cdot d y = 0$ This is not exact since calling $P (x, y) = - y^{2}, Q (x, y) = x^{2} + x y$ , we have $P_{y} = - 2 y \neq Q_{x} = 2 x + y$ .

To solve this type of equations we need to find $U (x, y)$ such that $U_{x} = P$ and $U_{y} = Q$ and the solution is given by $U (x y, y) = C$ , where it $C$ is a constant.

To solve this type of equations we will proceed as follows.

We have: $U_{x} (x, y) = P (x, y)$ . We integrate on both sides of the equality with respect to $x$ : $\int U_{x} (x, y) d x = \int P (x, y) d x \Rightarrow U (x, y) = \int P (x, y) d x + h (y)$

Therefore we have the function we are looking for, except for the fact that we do not know $h$ , a function that only depends on $y$ . To find it, we derive the previous expression with respect to $y$ : $U_{y} (x, y) = \frac{d}{d y} \int P (x, y) d x + h^{'} (y)$ Also, we know that $U_{y} = Q$ . Therefore equating terms we obtain a diifferentiable equation (that does not depend on $x$ , since the ODE is exact) and we can find $h (y)$ .

Once we know $h (y)$ , we add it to the expression of $U (x, y)$ which, equated to a constant, is the solution of our ODE.

Example

Let's solve the ODE $(x^{3} + y^{3}) d x + 3 x y^{2} d y = 0$ which we know to be exact.

We know that we are looking for a function $U (x, y)$ so that $U_{x} (x, y) = P (x, y)$ . As we have already seen we have: $U (x, y) = \int P (x, y) d x + h (y) = \int (x^{3} + y^{3}) d x + h (y) = \frac{x^{4}}{4} + y^{3} x + h (y)$ where $h (y)$ is a function to be determined that only depends on $y$ . To find this function, let's impose that $U$ is a solution, that is $U_{y} (x, y) = Q (x, y)$ .

Deriving with respect to $y$ we obtain a solution by equating it to a constant: $\begin{array}{r} U_{y} = 3 y^{2} x + h^{'} (y) \\ U_{y} = Q (x, y) = 3 x y^{2} \end{array}} \Rightarrow h^{'} (y) = 0 \Rightarrow h (y) = C$ where $C$ is a constant that is not so important since we will also have a constant in the final solution.

And so, the solution to the ODE will be: $U (x, y) = \frac{x^{4}}{4} + x y^{3} = C$