# Exact Equations

A first-order differential equation is an exact equation if it can be written in the form:

$$M(x,y)dx + N(x,y)dy = 0$$

Where $$M(x,y) = \cfrac{\partial f}{\partial x}$$ and $$N(x,y) = \cfrac{\partial f}{ \partial y}$$ and the expression on the left hand side is in fact an exact differential. $$M(x,y)$$ and $$N(x,y)$$ and their first partial derivatives must be continueous on some region $$\Re$$.

We can confirm the expression is an exact differential by checking

$$\cfrac{\partial M}{\partial y} = \cfrac{\partial N}{\partial x}$$

$$\cfrac{\partial}{\partial y}\cfrac{\partial f}{\partial x} = \cfrac{\partial }{\partial x}\cfrac{\partial f}{\partial y}$$

Confirm the expression $$(y \cos x + 3x^2 e^y) + (\sin x + x^3 e^y)y^{'}$$ is an exact differential.

The steps to solve an exact equation DE are:

1. Write the equation in the form $$M(x,y)dx + N(x,y)dy = 0$$.
2. Integrate $$M(x,y) = \frac{\partial f}{\partial x}$$ wrt $$x$$. Remember there will be a term $$h(y)$$.
3. Take the derivative wrt $$y$$ and set it equal to $$N(x,y) = \frac{\partial f}{\partial y}$$.
4. Solve for $$h(y)$$ by integrating.
5. Obtain an explicit solution for $$y$$ equal to a constant.

Find the solution to the DE $$(3x^2-2xy+2)dx + (6y^2-x^2+3)dy = 0$$.