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### Exact Equations

#### Additional Examples

Solve the following initial value problem,
$$\begin{align}
\frac{dy}{dx} &= \frac{-40xy + 8x + 4y + 4}{20x^2 - 4x + 21} \\
y(2) &= -1
\end{align}$$
This can be written as an exact equation. First we find the general solution following the paradigm.
$x = 2$ and $y = -1$ and solve for $K = -117$. So the solution to the initial value problem is
$$
20x^2y - 4x^2 - 4xy + 21y - 4x = -117
$$
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If you have any problems with this page, please contact bennett@math.ksu.edu.

©2010, 2014 Andrew G. Bennett

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- We write the equation in the standard form,
M dx + N dy = 0 . $$ (40xy - 8x - 4y - 4) dx + (20x^2 - 4x + 21) dy = 0 $$ - We test for exactness. $$\frac{\partial}{\partial y}\left(40xy - 8x - 4y - 4\right) = 40x - 4 = \frac{\partial}{\partial x}\left(20x^2 - 4x + 21\right) $$ so the equation is exact.
- Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= 40xy - 8x - 4y - 4\\ \frac{\partial F}{\partial y} &= 20x^2 - 4x + 21 \end{align}$$
- Integrate the first partial differential equation. $$ F(x,y) = \int (40xy - 8x - 4y - 4)\,\partial x = 20x^2y - 4x^2 - 4xy - 4x + C(y) $$
- Integrate the second partial differential equation. $$ F(x,y) = \int (20x^2 - 4x + 21)\,\partial y = 20x^2y - 4xy + 21y + \tilde{C}(x) $$
- Equate the expressions for F(x,y).
Matching the expressions up, we find $C(y) = 21y$ and $ \tilde{C}(x) = -4x^2 - 4x. $ So $$ F(x,y) = 20x^2y - 4x^2 - 4xy + 21y - 4x. $$

- The solution is $F(x,y) = K.$ $$ 20x^2y - 4x^2 - 4xy + 21y - 4x = K $$

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

©2010, 2014 Andrew G. Bennett