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, $$\begin{align} \frac{dy}{dx} &= \frac{10xy + 6x + 5y + 3}{-5x^2 - 12y^2 - 4y - 5x + 6} \\ y(-3) &= 1 \end{align}$$ 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. $$ (-10xy - 6x - 5y - 3) dx + (-5x^2 - 12y^2 - 4y - 5x + 6) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(-10xy - 6x - 5y - 3\right) = -10x - 5 = \frac{\partial}{\partial x}\left(-5x^2 - 12y^2 - 4y - 5x + 6\right) $$ so the equation is exact.

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

    Matching the expressions up, we find $C(y) = -4y^3 - 2y^2 + 6y$ and $ \tilde{C}(x) = -3x^2 - 3x. $ So $$ F(x,y) = -4y^3 - 5x^2y - 3x^2 - 5xy - 2y^2 + 6y - 3x. $$

  7. The solution is $F(x,y) = K.$ $$ -4y^3 - 5x^2y - 3x^2 - 5xy - 2y^2 + 6y - 3x = K $$
Now we plug in the initial values $x = -3$ and $y = 1$ and solve for $K = -48$. So the solution to the initial value problem is $$ -4y^3 - 5x^2y - 3x^2 - 5xy - 2y^2 + 6y - 3x = -48 $$ You may reload this page to generate additional examples.


If you have any problems with this page, please contact bennett@math.ksu.edu.
©2010, 2014 Andrew G. Bennett