# How to Solve Exact Differential Equations

fact
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 $$\Phi_y = N$$
fact
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 $$\Phi (x, y)$$
fact
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: $$M_y = N_x$$
derivation
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.
fact
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.
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.
example
practice problems