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, $$\begin{align} \frac{dy}{dx} &= \frac{20y^2 + 15y - 6x - 6}{9y^2 - 40xy - 44y - 15x - 14} \\ y(-2) &= -1 \end{align}$$ 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. $$ (-20y^2 - 15y + 6x + 6) dx + (9y^2 - 40xy - 44y - 15x - 14) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(-20y^2 - 15y + 6x + 6\right) = -40y - 15 = \frac{\partial}{\partial x}\left(9y^2 - 40xy - 44y - 15x - 14\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= -20y^2 - 15y + 6x + 6\\ \frac{\partial F}{\partial y} &= 9y^2 - 40xy - 44y - 15x - 14 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (-20y^2 - 15y + 6x + 6)\,\partial x = -20xy^2 - 15xy + 3x^2 + 6x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (9y^2 - 40xy - 44y - 15x - 14)\,\partial y = 3y^3 - 20xy^2 - 22y^2 - 15xy - 14y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = 3y^3 - 22y^2 - 14y$ and $ \tilde{C}(x) = 3x^2 + 6x. $ So $$ F(x,y) = 3y^3 - 20xy^2 - 15xy + 3x^2 - 22y^2 + 6x - 14y. $$

  7. The solution is $F(x,y) = K.$ $$ 3y^3 - 20xy^2 - 15xy + 3x^2 - 22y^2 + 6x - 14y = K $$
Now we plug in the initial values $x = -2$ and $y = -1$ and solve for $K = -1$. So the solution to the initial value problem is $$ 3y^3 - 20xy^2 - 15xy + 3x^2 - 22y^2 + 6x - 14y = -1 $$ 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