Exact Equations
Additional Examples
- We write the equation in the standard form,
M dx + N dy = 0 . $$ (8xy + 32x - y - 7) dx + (6y^2 + 4x^2 - 4y - x - 1) dy = 0 $$ - We test for exactness. $$\frac{\partial}{\partial y}\left(8xy + 32x - y - 7\right) = 8x - 1 = \frac{\partial}{\partial x}\left(6y^2 + 4x^2 - 4y - x - 1\right) $$ so the equation is exact.
- Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= 8xy + 32x - y - 7\\ \frac{\partial F}{\partial y} &= 6y^2 + 4x^2 - 4y - x - 1 \end{align}$$
- Integrate the first partial differential equation. $$ F(x,y) = \int (8xy + 32x - y - 7)\,\partial x = 4x^2y + 16x^2 - xy - 7x + C(y) $$
- Integrate the second partial differential equation. $$ F(x,y) = \int (6y^2 + 4x^2 - 4y - x - 1)\,\partial y = 2y^3 + 4x^2y - 2y^2 - xy - y + \tilde{C}(x) $$
- Equate the expressions for F(x,y). Matching the expressions up, we find $C(y) = 2y^3 - 2y^2 - y$ and $ \tilde{C}(x) = 16x^2 - 7x. $ So $$ F(x,y) = 2y^3 + 4x^2y - xy + 16x^2 - 2y^2 - 7x - y. $$
- The solution is $F(x,y) = K.$ $$ 2y^3 + 4x^2y - xy + 16x^2 - 2y^2 - 7x - y = K $$
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