Exact differential equation

An exact differential equation is a differential equation that can be solved in the following manner.

Suppose you are given an equation of the form:

<math>M(t,y) + N(t,y)y' = 0\,</math> or <math>M(t,y) dt + N(t,y) dy = 0\,</math>

To find the solution of this equation, we assume that the solution is φ = constant, where φ is determined by integrating M and N.

<math>\phi(t, y) = \int_0^t M(s, 0) ds + \int_0^y N(t, s) ds</math>

Go through the example to find φ by integrating, then check that

<math>\frac{\partial \phi}{\partial t} = M</math>

and

<math>\frac{\partial \phi}{\partial y} = N</math>

and that any function φ = some constant, when turned into the corresponding dy/dt, satisfies the original equation. Be sure to emphasize that one must check first that

<math>\frac{\partial M}{\partial y} = \frac{\partial N}{\partial t}</math>

(That's the condition for "exactness" of the differential form M dt + N dy.)

where <math>\frac{\partial M}{\partial y} = \frac{\partial N}{\partial t}</math>.

To find <math>y</math>, first set <math>M = \frac{\partial \phi}{\partial t}</math> and <math>N = \frac{\partial \phi}{\partial y}</math>. Then manipulate to get <math>M \partial t = \partial \phi</math> and <math>N \partial y = \partial \phi</math>. Integrate both sides, compare the results for <math>\phi</math>, and combine the terms into one equation (for terms that show up in both expressions, only write once in the combined expression.) To solve the expression for <math>y</math>, plug into the quadratic formula.