Exact Equations
Additional Examples
- We write the equation in the standard form,
M dx + N dy = 0 . $$ (-18xy + 30x + 12y - 18) dx + (-9x^2 + 12x + 10y - 9) dy = 0 $$ - We test for exactness. $$\frac{\partial}{\partial y}\left(-18xy + 30x + 12y - 18\right) = -18x + 12 = \frac{\partial}{\partial x}\left(-9x^2 + 12x + 10y - 9\right) $$ so the equation is exact.
- Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= -18xy + 30x + 12y - 18\\ \frac{\partial F}{\partial y} &= -9x^2 + 12x + 10y - 9 \end{align}$$
- Integrate the first partial differential equation. $$ F(x,y) = \int (-18xy + 30x + 12y - 18)\,\partial x = -9x^2y + 15x^2 + 12xy - 18x + C(y) $$
- Integrate the second partial differential equation. $$ F(x,y) = \int (-9x^2 + 12x + 10y - 9)\,\partial y = -9x^2y + 12xy + 5y^2 - 9y + \tilde{C}(x) $$
- Equate the expressions for F(x,y). Matching the expressions up, we find $C(y) = 5y^2 - 9y$ and $ \tilde{C}(x) = 15x^2 - 18x. $ So $$ F(x,y) = -9x^2y + 12xy + 15x^2 + 5y^2 - 18x - 9y. $$
- The solution is $F(x,y) = K.$ $$ -9x^2y + 12xy + 15x^2 + 5y^2 - 18x - 9y = K $$
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