Exact Form

If a first order equation isn’t separable or linear, and it can’t be solved using substitutions, then it can be solved if it is in exact form. If it isn’t in exact form, it can be put in exact form.

The Essentials

Every first order differential equation fits in this form:

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

The equation is in exact form if:

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

Or, using a different notation:

\[ M_{y} = N_{x} \]

If the equation is in exact form, there exists a potential function \( \phi(x, y)\) where \( \phi_{x} = M \) and \( \phi_{y} = N \). To find the potential function, either do the ’partial integral’ of M or N: \( \int M dx \) or \( \int N dy \). If we choose M, we integrate and instead of adding a constant of integration, C, we add a function of y, \( h(y) \). Then we can take the partial derivative of the result with respect to y and match the \( h′(y) \) term with the appropriate term in N. Likewise, if we choose N, our constant of integration becomes a function of x, \( f(x) \), and we take the partial derivative with respect to x. After we have the potential function, we set it equal to zero and solve for \( y(x) \). The equation will often not be able to be solved for \( y(x) \) in which cases it can be left implicit.

Example

Solve the equation:

\[ 3x^2y + (x^3 - 1) \frac{dy}{dx} = 0 \]

This equation is already in standard form with:

\[ M(x, y) = 3x^2y \] \[ N(x, y) = x^3 - 1 \]

First, we text for exactness:

\[ \frac{\partial M}{\partial y} = 3x^2 \] \[ \frac{\partial N}{\partial x} = 3x^2 \]

Since the partial of M with respect to y and the partial of N with respect to x are equal, the equation is in exact form and we can find the potential function. Next, we integrate. We can choose either M or N, whichever we would rather integrate. In this case we use M. Since \( \phi_x = M \), we integrate with respect to x:

\[ \phi(x, y) = \int M \cdot dx \] \[ \phi(x, y) = \int 3x^2y \cdot dx \] \[ \phi(x, y) = x^3y + h(y) \]

Since we treated y as a constant, are constant of integration is a function of y. Now, since \( \phi_y = N \) we can take the partial derivative with respect to y and the potential function should be equal to N to find \( h(y) \):

\[ \phi_y = N \] \[ x^3 + h'(y) = x^3 - 1 \]

Here, clearly \( h'(y) = -1 \). Therefore \( h(y) = -y + C \). Now we have the potential function:

\[ \phi(x, y) = x^3y - y + C \]

Next, we set the potential function equal to zero and solve for y:

\[ x^3y - y = C \] \[ y(x) = \frac{C}{x^3 - 1} \]

Practice

Verify that this equation is exact, find the potential function, and solve the equation:

\[ 2xy + y^2 + (2xy + x^2) \frac{dy}{dx} = 0 \]

Solution:

Potential function: \( \phi(x, y) = x^2y+xy^2 + C \). Solution: \( x^2y + xy^2 = C \)