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, $$\begin{align} \frac{dy}{dx} &= \frac{10y^2 + 20y + 4x + 19}{-20xy - 20x + 14y + 5} \\ y(0) &= -1 \end{align}$$ 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. $$ (-10y^2 - 20y - 4x - 19) dx + (-20xy - 20x + 14y + 5) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(-10y^2 - 20y - 4x - 19\right) = -20y - 20 = \frac{\partial}{\partial x}\left(-20xy - 20x + 14y + 5\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= -10y^2 - 20y - 4x - 19\\ \frac{\partial F}{\partial y} &= -20xy - 20x + 14y + 5 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (-10y^2 - 20y - 4x - 19)\,\partial x = -10xy^2 - 20xy - 2x^2 - 19x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (-20xy - 20x + 14y + 5)\,\partial y = -10xy^2 - 20xy + 7y^2 + 5y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = 7y^2 + 5y$ and $ \tilde{C}(x) = -2x^2 - 19x. $ So $$ F(x,y) = -10xy^2 - 20xy - 2x^2 + 7y^2 - 19x + 5y. $$

  7. The solution is $F(x,y) = K.$ $$ -10xy^2 - 20xy - 2x^2 + 7y^2 - 19x + 5y = K $$
Now we plug in the initial values $x = 0$ and $y = -1$ and solve for $K = 2$. So the solution to the initial value problem is $$ -10xy^2 - 20xy - 2x^2 + 7y^2 - 19x + 5y = 2 $$ 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