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, $$\begin{align} \frac{dy}{dx} &= \frac{16xy + 32x + 4y + 7}{-8x^2 - 4x - 4y - 8} \\ y(2) &= -3 \end{align}$$ 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. $$ (-16xy - 32x - 4y - 7) dx + (-8x^2 - 4x - 4y - 8) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(-16xy - 32x - 4y - 7\right) = -16x - 4 = \frac{\partial}{\partial x}\left(-8x^2 - 4x - 4y - 8\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= -16xy - 32x - 4y - 7\\ \frac{\partial F}{\partial y} &= -8x^2 - 4x - 4y - 8 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (-16xy - 32x - 4y - 7)\,\partial x = -8x^2y - 16x^2 - 4xy - 7x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (-8x^2 - 4x - 4y - 8)\,\partial y = -8x^2y - 4xy - 2y^2 - 8y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = -2y^2 - 8y$ and $ \tilde{C}(x) = -16x^2 - 7x. $ So $$ F(x,y) = -8x^2y - 4xy - 16x^2 - 2y^2 - 7x - 8y. $$

  7. The solution is $F(x,y) = K.$ $$ -8x^2y - 4xy - 16x^2 - 2y^2 - 7x - 8y = K $$
Now we plug in the initial values $x = 2$ and $y = -3$ and solve for $K = 48$. So the solution to the initial value problem is $$ -8x^2y - 4xy - 16x^2 - 2y^2 - 7x - 8y = 48 $$ 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