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.

Show Answer

The steps to solve an exact equation DE are:

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

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

Show Answer