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, $$\begin{align} \frac{dy}{dx} &= \frac{16y^2 + 20y - 4x - 5}{-32xy - 20x - 8y - 1} \\ y(-2) &= 3 \end{align}$$ 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. $$ (-16y^2 - 20y + 4x + 5) dx + (-32xy - 20x - 8y - 1) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(-16y^2 - 20y + 4x + 5\right) = -32y - 20 = \frac{\partial}{\partial x}\left(-32xy - 20x - 8y - 1\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= -16y^2 - 20y + 4x + 5\\ \frac{\partial F}{\partial y} &= -32xy - 20x - 8y - 1 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (-16y^2 - 20y + 4x + 5)\,\partial x = -16xy^2 - 20xy + 2x^2 + 5x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (-32xy - 20x - 8y - 1)\,\partial y = -16xy^2 - 20xy - 4y^2 - y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = -4y^2 - y$ and $ \tilde{C}(x) = 2x^2 + 5x. $ So $$ F(x,y) = -16xy^2 - 20xy + 2x^2 - 4y^2 + 5x - y. $$

  7. The solution is $F(x,y) = K.$ $$ -16xy^2 - 20xy + 2x^2 - 4y^2 + 5x - y = K $$
Now we plug in the initial values $x = -2$ and $y = 3$ and solve for $K = 367$. So the solution to the initial value problem is $$ -16xy^2 - 20xy + 2x^2 - 4y^2 + 5x - y = 367 $$ 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