Kansas State University

Mathematics Department

Textbook Contents

Warning: MathJax requires JavaScript to process the mathematics on this page.
If your browser supports JavaScript, be sure it is enabled.

Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{16y - 5}{-16x + 11} \\ y(-3) &= 4 \end{align}$$ This can be written as an exact equation. First we find the general solution following the paradigm.

  1. We write the equation in the standard form, M dx + N dy = 0. $$ (-16y + 5) dx + (-16x + 11) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(-16y + 5\right) = -16 = \frac{\partial}{\partial x}\left(-16x + 11\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= -16y + 5\\ \frac{\partial F}{\partial y} &= -16x + 11 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (-16y + 5)\,\partial x = -16xy + 5x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (-16x + 11)\,\partial y = -16xy + 11y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = 11y$ and $ \tilde{C}(x) = 5x. $ So $$ F(x,y) = -16xy + 5x + 11y. $$

  7. The solution is $F(x,y) = K.$ $$ -16xy + 5x + 11y = K $$
Now we plug in the initial values $x = -3$ and $y = 4$ and solve for $K = 221$. So the solution to the initial value problem is $$ -16xy + 5x + 11y = 221 $$ You may reload this page to generate additional examples.

If you have any problems with this page, please contact bennett@ksu.edu.
©1994-2025 Andrew G. Bennett