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, $$\begin{align} \frac{dy}{dx} &= \frac{-3y^2 - 4y + 8x + 1}{6xy + 4x - 16y - 7} \\ y(-2) &= 0 \end{align}$$ 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. $$ (3y^2 + 4y - 8x - 1) dx + (6xy + 4x - 16y - 7) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(3y^2 + 4y - 8x - 1\right) = 6y + 4 = \frac{\partial}{\partial x}\left(6xy + 4x - 16y - 7\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= 3y^2 + 4y - 8x - 1\\ \frac{\partial F}{\partial y} &= 6xy + 4x - 16y - 7 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (3y^2 + 4y - 8x - 1)\,\partial x = 3xy^2 + 4xy - 4x^2 - x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (6xy + 4x - 16y - 7)\,\partial y = 3xy^2 + 4xy - 8y^2 - 7y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = -8y^2 - 7y$ and $ \tilde{C}(x) = -4x^2 - x. $ So $$ F(x,y) = 3xy^2 + 4xy - 4x^2 - 8y^2 - x - 7y. $$

  7. The solution is $F(x,y) = K.$ $$ 3xy^2 + 4xy - 4x^2 - 8y^2 - x - 7y = K $$
Now we plug in the initial values $x = -2$ and $y = 0$ and solve for $K = -14$. So the solution to the initial value problem is $$ 3xy^2 + 4xy - 4x^2 - 8y^2 - x - 7y = -14 $$ 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