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,
$$\begin{align}
\frac{dy}{dx} &= \frac{4y - 12}{3y^2 - 4x - 2y + 9} \\
y(1) &= -2
\end{align}$$
This can be written as an exact equation. First we find the general solution following the paradigm.
$x = 1$ and $y = -2$ and solve for $K = -10$. So the solution to the initial value problem is
$$
y^3 - 4xy - y^2 + 9y + 12x = -10
$$
*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

If your browser supports JavaScript, be sure it is enabled.

- We write the equation in the standard form,
M dx + N dy = 0 . $$ (-4y + 12) dx + (3y^2 - 4x - 2y + 9) dy = 0 $$ - We test for exactness. $$\frac{\partial}{\partial y}\left(-4y + 12\right) = -4 = \frac{\partial}{\partial x}\left(3y^2 - 4x - 2y + 9\right) $$ so the equation is exact.
- Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= -4y + 12\\ \frac{\partial F}{\partial y} &= 3y^2 - 4x - 2y + 9 \end{align}$$
- Integrate the first partial differential equation. $$ F(x,y) = \int (-4y + 12)\,\partial x = -4xy + 12x + C(y) $$
- Integrate the second partial differential equation. $$ F(x,y) = \int (3y^2 - 4x - 2y + 9)\,\partial y = y^3 - 4xy - y^2 + 9y + \tilde{C}(x) $$
- Equate the expressions for F(x,y).
Matching the expressions up, we find $C(y) = y^3 - y^2 + 9y$ and $ \tilde{C}(x) = 12x. $ So $$ F(x,y) = y^3 - 4xy - y^2 + 9y + 12x. $$

- The solution is $F(x,y) = K.$ $$ y^3 - 4xy - y^2 + 9y + 12x = K $$

If you have any problems with this page, please contact bennett@math.ksu.edu.

©2010, 2014 Andrew G. Bennett