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, $$\begin{align} \frac{dy}{dx} &= \frac{-25y^2 - 5y + 11}{50xy + 5x - 40y - 3} \\ y(0) &= -2 \end{align}$$ 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. $$ (25y^2 + 5y - 11) dx + (50xy + 5x - 40y - 3) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(25y^2 + 5y - 11\right) = 50y + 5 = \frac{\partial}{\partial x}\left(50xy + 5x - 40y - 3\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= 25y^2 + 5y - 11\\ \frac{\partial F}{\partial y} &= 50xy + 5x - 40y - 3 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (25y^2 + 5y - 11)\,\partial x = 25xy^2 + 5xy - 11x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (50xy + 5x - 40y - 3)\,\partial y = 25xy^2 + 5xy - 20y^2 - 3y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = -20y^2 - 3y$ and $ \tilde{C}(x) = -11x. $ So $$ F(x,y) = 25xy^2 + 5xy - 20y^2 - 11x - 3y. $$

  7. The solution is $F(x,y) = K.$ $$ 25xy^2 + 5xy - 20y^2 - 11x - 3y = K $$
Now we plug in the initial values $x = 0$ and $y = -2$ and solve for $K = -74$. So the solution to the initial value problem is $$ 25xy^2 + 5xy - 20y^2 - 11x - 3y = -74 $$ 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