Exact Equations
Additional Examples
- We write the equation in the standard form,
M dx + N dy = 0 . $$ (20y - 14) dx + (15y^2 + 2y + 20x - 24) dy = 0 $$ - We test for exactness. $$\frac{\partial}{\partial y}\left(20y - 14\right) = 20 = \frac{\partial}{\partial x}\left(15y^2 + 2y + 20x - 24\right) $$ so the equation is exact.
- Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= 20y - 14\\ \frac{\partial F}{\partial y} &= 15y^2 + 2y + 20x - 24 \end{align}$$
- Integrate the first partial differential equation. $$ F(x,y) = \int (20y - 14)\,\partial x = 20xy - 14x + C(y) $$
- Integrate the second partial differential equation. $$ F(x,y) = \int (15y^2 + 2y + 20x - 24)\,\partial y = 5y^3 + y^2 + 20xy - 24y + \tilde{C}(x) $$
- Equate the expressions for F(x,y). Matching the expressions up, we find $C(y) = 5y^3 + y^2 - 24y$ and $ \tilde{C}(x) = -14x. $ So $$ F(x,y) = 5y^3 + 20xy + y^2 - 14x - 24y. $$
- The solution is $F(x,y) = K.$ $$ 5y^3 + 20xy + y^2 - 14x - 24y = K $$
If you have any problems with this page, please contact bennett@math.ksu.edu.
©2010, 2014 Andrew G. Bennett