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, $$\begin{align} \frac{dy}{dx} &= \frac{-2y^2 + 10y - 2x + 9}{-9y^2 + 4xy - 10x + 10} \\ y(-3) &= -4 \end{align}$$ 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. $$ (2y^2 - 10y + 2x - 9) dx + (-9y^2 + 4xy - 10x + 10) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(2y^2 - 10y + 2x - 9\right) = 4y - 10 = \frac{\partial}{\partial x}\left(-9y^2 + 4xy - 10x + 10\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= 2y^2 - 10y + 2x - 9\\ \frac{\partial F}{\partial y} &= -9y^2 + 4xy - 10x + 10 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (2y^2 - 10y + 2x - 9)\,\partial x = 2xy^2 - 10xy + x^2 - 9x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (-9y^2 + 4xy - 10x + 10)\,\partial y = -3y^3 + 2xy^2 - 10xy + 10y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = -3y^3 + 10y$ and $ \tilde{C}(x) = x^2 - 9x. $ So $$ F(x,y) = 2xy^2 - 3y^3 - 10xy + x^2 - 9x + 10y. $$

  7. The solution is $F(x,y) = K.$ $$ 2xy^2 - 3y^3 - 10xy + x^2 - 9x + 10y = K $$
Now we plug in the initial values $x = -3$ and $y = -4$ and solve for $K = -28$. So the solution to the initial value problem is $$ 2xy^2 - 3y^3 - 10xy + x^2 - 9x + 10y = -28 $$ 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