Math 240 Home, Textbook Contents, Online Homework Home

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, $$\begin{align} \frac{dy}{dx} &= \frac{-24xy + 26x - 3y + 6}{12x^2 + 3y^2 + 4y + 3x + 4} \\ y(-2) &= 0 \end{align}$$ 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. $$ (24xy - 26x + 3y - 6) dx + (12x^2 + 3y^2 + 4y + 3x + 4) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(24xy - 26x + 3y - 6\right) = 24x + 3 = \frac{\partial}{\partial x}\left(12x^2 + 3y^2 + 4y + 3x + 4\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= 24xy - 26x + 3y - 6\\ \frac{\partial F}{\partial y} &= 12x^2 + 3y^2 + 4y + 3x + 4 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (24xy - 26x + 3y - 6)\,\partial x = 12x^2y - 13x^2 + 3xy - 6x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (12x^2 + 3y^2 + 4y + 3x + 4)\,\partial y = 12x^2y + y^3 + 2y^2 + 3xy + 4y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = y^3 + 2y^2 + 4y$ and $ \tilde{C}(x) = -13x^2 - 6x. $ So $$ F(x,y) = y^3 + 12x^2y - 13x^2 + 3xy + 2y^2 + 4y - 6x. $$

  7. The solution is $F(x,y) = K.$ $$ y^3 + 12x^2y - 13x^2 + 3xy + 2y^2 + 4y - 6x = K $$
Now we plug in the initial values $x = -2$ and $y = 0$ and solve for $K = -40$. So the solution to the initial value problem is $$ y^3 + 12x^2y - 13x^2 + 3xy + 2y^2 + 4y - 6x = -40 $$ You may reload this page to generate additional examples.


If you have any problems with this page, please contact bennett@math.ksu.edu.
©2010 Andrew G. Bennett