Exact Equations
Additional Examples
- We write the equation in the standard form,
M dx + N dy = 0 . $$ (-8x + 10y - 1) dx + (10x + 4y + 6) dy = 0 $$ - We test for exactness. $$\frac{\partial}{\partial y}\left(-8x + 10y - 1\right) = 10 = \frac{\partial}{\partial x}\left(10x + 4y + 6\right) $$ so the equation is exact.
- Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= -8x + 10y - 1\\ \frac{\partial F}{\partial y} &= 10x + 4y + 6 \end{align}$$
- Integrate the first partial differential equation. $$ F(x,y) = \int (-8x + 10y - 1)\,\partial x = -4x^2 + 10xy - x + C(y) $$
- Integrate the second partial differential equation. $$ F(x,y) = \int (10x + 4y + 6)\,\partial y = 10xy + 2y^2 + 6y + \tilde{C}(x) $$
- Equate the expressions for F(x,y). Matching the expressions up, we find $C(y) = 2y^2 + 6y$ and $ \tilde{C}(x) = -4x^2 - x. $ So $$ F(x,y) = -4x^2 + 10xy + 2y^2 - x + 6y. $$
- The solution is $F(x,y) = K.$ $$ -4x^2 + 10xy + 2y^2 - x + 6y = K $$
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