How to Solve Exact Differential Equations

An exact differential equation is one which can be written as: $$ M(x, y) + N(x, y)\frac{dy}{dx} = 0 $$ Where there is some function \(\Phi (x, y)\) such that: $$ \Phi_x = M $$ And $$ \Phi_y = N $$

The solution to an exact differential equation is given by: $$ \Phi (x, y) = c $$ For some constant \(c\). So solving an exact differential equation amounts to finding the function \(\Phi (x, y)\)

To check whether a differential equation is exact place it in the form \(M(x, y) + N(x, y)\frac{dy}{dx} = 0\) and then check that: $$ M_y = N_x $$

We know that in an exact differential equation \(M = \Phi_x\) and \(N = \Phi_y\). Assuming \(\Phi\) and its derivatives are continuous then we also know that \(\Phi_{xy} = \Phi_{yx}\) From this we can see that \(M_y = N_x\). This is a necessary and sufficient condition for the equation to be exact.

The process of solving an exact differential equation for \(\Phi\) is long but not tricky. The difficulty comes from performing a large number of steps without error rather than any one step being hard. You already have all the skills necessary, we just need to put them together into a strategy.

1. Integrate \(M(x, y)dx\) to find an expression for \(\Phi\) that will include some unknown function \(h(y)\) instead of a constant \(c\) since expressions of just \(y\) will disappeared when the \(x\) derivative is taken.
2. Take the derivative of our \(\Phi\) we just found and equate it to \(N(x, y)\) to find what the derivative of our unknown \(h(y)\) is.
3. Integrate \(h'(y)\) to find \(h(y)\) up to an additive constant \(c\) to complete our solution for \(\Phi\)

We could also integrate \(N\) first and get an \(h(x)\) unknown expression however to keep things simpler we'll tend to integrate \(M\). This is only to make the process easier to learn. The entire process is more easily describe through example.

Solve \(2xy + x^2\frac{dy}{dx} = 0\)

First we check that the equation is exact. \(M = 2xy\) and \(N = x^2\) \(M_y = 2x = N_x\) so the equation is exact. Now we'll find an expression for \(\Phi\): \(\Phi (x, y) = \int Mdx = \int 2xydx = x^2y + h(y)\) Now \(\Phi_y = x^2 + h'(y) = N = x^2 \implies h'(y) = 0\) Integrating \(h'(y)\) we find that: \(h(y) = \int h'(y)dy = \int 0dy = c\) So putting it all together we have: \(\Phi = x^2y + c\) And our solution is therefore \(x^2y = c\) where this \(c\) is a different constant from the previous one. We've reused the symbol because having 9 \(c_x\)'s around the place is confusing when unnecessary.