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, $$\begin{align} \frac{dy}{dx} &= \frac{6y^2 - 6y - 10}{-12xy + 6x - 6y + 9} \\ y(2) &= -5 \end{align}$$ 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. $$ (-6y^2 + 6y + 10) dx + (-12xy + 6x - 6y + 9) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(-6y^2 + 6y + 10\right) = -12y + 6 = \frac{\partial}{\partial x}\left(-12xy + 6x - 6y + 9\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= -6y^2 + 6y + 10\\ \frac{\partial F}{\partial y} &= -12xy + 6x - 6y + 9 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (-6y^2 + 6y + 10)\,\partial x = -6xy^2 + 6xy + 10x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (-12xy + 6x - 6y + 9)\,\partial y = -6xy^2 + 6xy - 3y^2 + 9y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = -3y^2 + 9y$ and $ \tilde{C}(x) = 10x. $ So $$ F(x,y) = -6xy^2 + 6xy - 3y^2 + 10x + 9y. $$

  7. The solution is $F(x,y) = K.$ $$ -6xy^2 + 6xy - 3y^2 + 10x + 9y = K $$
Now we plug in the initial values $x = 2$ and $y = -5$ and solve for $K = -460$. So the solution to the initial value problem is $$ -6xy^2 + 6xy - 3y^2 + 10x + 9y = -460 $$ 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