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, $$\begin{align} \frac{dy}{dx} &= \frac{-32xy - 48x + 8y + 5}{16x^2 - 8x - 2y - 5} \\ y(0) &= -1 \end{align}$$ 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. $$ (32xy + 48x - 8y - 5) dx + (16x^2 - 8x - 2y - 5) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(32xy + 48x - 8y - 5\right) = 32x - 8 = \frac{\partial}{\partial x}\left(16x^2 - 8x - 2y - 5\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= 32xy + 48x - 8y - 5\\ \frac{\partial F}{\partial y} &= 16x^2 - 8x - 2y - 5 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (32xy + 48x - 8y - 5)\,\partial x = 16x^2y + 24x^2 - 8xy - 5x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (16x^2 - 8x - 2y - 5)\,\partial y = 16x^2y - 8xy - y^2 - 5y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = -y^2 - 5y$ and $ \tilde{C}(x) = 24x^2 - 5x. $ So $$ F(x,y) = 16x^2y - 8xy + 24x^2 - y^2 - 5x - 5y. $$

  7. The solution is $F(x,y) = K.$ $$ 16x^2y - 8xy + 24x^2 - y^2 - 5x - 5y = K $$
Now we plug in the initial values $x = 0$ and $y = -1$ and solve for $K = 4$. So the solution to the initial value problem is $$ 16x^2y - 8xy + 24x^2 - y^2 - 5x - 5y = 4 $$ 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