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, $$\begin{align} \frac{dy}{dx} &= \frac{-40xy + 32x + 20y - 12}{-6y^2 + 20x^2 - 2y - 20x + 21} \\ y(1) &= -5 \end{align}$$ 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. $$ (40xy - 32x - 20y + 12) dx + (-6y^2 + 20x^2 - 2y - 20x + 21) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(40xy - 32x - 20y + 12\right) = 40x - 20 = \frac{\partial}{\partial x}\left(-6y^2 + 20x^2 - 2y - 20x + 21\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= 40xy - 32x - 20y + 12\\ \frac{\partial F}{\partial y} &= -6y^2 + 20x^2 - 2y - 20x + 21 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (40xy - 32x - 20y + 12)\,\partial x = 20x^2y - 16x^2 - 20xy + 12x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (-6y^2 + 20x^2 - 2y - 20x + 21)\,\partial y = -2y^3 + 20x^2y - y^2 - 20xy + 21y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = -2y^3 - y^2 + 21y$ and $ \tilde{C}(x) = -16x^2 + 12x. $ So $$ F(x,y) = -2y^3 + 20x^2y - 20xy - 16x^2 - y^2 + 12x + 21y. $$

  7. The solution is $F(x,y) = K.$ $$ -2y^3 + 20x^2y - 20xy - 16x^2 - y^2 + 12x + 21y = K $$
Now we plug in the initial values $x = 1$ and $y = -5$ and solve for $K = 116$. So the solution to the initial value problem is $$ -2y^3 + 20x^2y - 20xy - 16x^2 - y^2 + 12x + 21y = 116 $$ 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