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, $$\begin{align} \frac{dy}{dx} &= \frac{-24xy + 20y + 8x}{15y^2 + 12x^2 + 4y - 20x + 8} \\ y(2) &= 0 \end{align}$$ 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. $$ (24xy - 20y - 8x) dx + (15y^2 + 12x^2 + 4y - 20x + 8) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(24xy - 20y - 8x\right) = 24x - 20 = \frac{\partial}{\partial x}\left(15y^2 + 12x^2 + 4y - 20x + 8\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= 24xy - 20y - 8x\\ \frac{\partial F}{\partial y} &= 15y^2 + 12x^2 + 4y - 20x + 8 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (24xy - 20y - 8x)\,\partial x = 12x^2y - 20xy - 4x^2 + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (15y^2 + 12x^2 + 4y - 20x + 8)\,\partial y = 5y^3 + 12x^2y + 2y^2 - 20xy + 8y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = 5y^3 + 2y^2 + 8y$ and $ \tilde{C}(x) = -4x^2. $ So $$ F(x,y) = 12x^2y + 5y^3 - 20xy - 4x^2 + 2y^2 + 8y. $$

  7. The solution is $F(x,y) = K.$ $$ 12x^2y + 5y^3 - 20xy - 4x^2 + 2y^2 + 8y = K $$
Now we plug in the initial values $x = 2$ and $y = 0$ and solve for $K = -16$. So the solution to the initial value problem is $$ 12x^2y + 5y^3 - 20xy - 4x^2 + 2y^2 + 8y = -16 $$ 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