Exact Equations
Additional Examples
- We write the equation in the standard form,
M dx + N dy = 0 . $$ (10y^2 - 4y - 10x + 5) dx + (15y^2 + 20xy - 30y - 4x - 1) dy = 0 $$ - We test for exactness. $$\frac{\partial}{\partial y}\left(10y^2 - 4y - 10x + 5\right) = 20y - 4 = \frac{\partial}{\partial x}\left(15y^2 + 20xy - 30y - 4x - 1\right) $$ so the equation is exact.
- Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= 10y^2 - 4y - 10x + 5\\ \frac{\partial F}{\partial y} &= 15y^2 + 20xy - 30y - 4x - 1 \end{align}$$
- Integrate the first partial differential equation. $$ F(x,y) = \int (10y^2 - 4y - 10x + 5)\,\partial x = 10xy^2 - 4xy - 5x^2 + 5x + C(y) $$
- Integrate the second partial differential equation. $$ F(x,y) = \int (15y^2 + 20xy - 30y - 4x - 1)\,\partial y = 5y^3 + 10xy^2 - 15y^2 - 4xy - y + \tilde{C}(x) $$
- Equate the expressions for F(x,y). Matching the expressions up, we find $C(y) = 5y^3 - 15y^2 - y$ and $ \tilde{C}(x) = -5x^2 + 5x. $ So $$ F(x,y) = 5y^3 + 10xy^2 - 4xy - 5x^2 - 15y^2 + 5x - y. $$
- The solution is $F(x,y) = K.$ $$ 5y^3 + 10xy^2 - 4xy - 5x^2 - 15y^2 + 5x - y = K $$
If you have any problems with this page, please contact bennett@ksu.edu.
©1994-2025 Andrew G. Bennett