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, $$\begin{align} \frac{dy}{dx} &= \frac{-30xy - 22x + 20y + 4}{15x^2 - 20x + 8y + 10} \\ y(-2) &= 1 \end{align}$$ 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. $$ (30xy + 22x - 20y - 4) dx + (15x^2 - 20x + 8y + 10) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(30xy + 22x - 20y - 4\right) = 30x - 20 = \frac{\partial}{\partial x}\left(15x^2 - 20x + 8y + 10\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= 30xy + 22x - 20y - 4\\ \frac{\partial F}{\partial y} &= 15x^2 - 20x + 8y + 10 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (30xy + 22x - 20y - 4)\,\partial x = 15x^2y + 11x^2 - 20xy - 4x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (15x^2 - 20x + 8y + 10)\,\partial y = 15x^2y - 20xy + 4y^2 + 10y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = 4y^2 + 10y$ and $ \tilde{C}(x) = 11x^2 - 4x. $ So $$ F(x,y) = 15x^2y - 20xy + 11x^2 + 4y^2 - 4x + 10y. $$

  7. The solution is $F(x,y) = K.$ $$ 15x^2y - 20xy + 11x^2 + 4y^2 - 4x + 10y = K $$
Now we plug in the initial values $x = -2$ and $y = 1$ and solve for $K = 166$. So the solution to the initial value problem is $$ 15x^2y - 20xy + 11x^2 + 4y^2 - 4x + 10y = 166 $$ 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