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### Exact Equations

Solve the following initial value problem, \begin{align} \frac{dy}{dx} &= \frac{-5y^2 - 6x + 2y - 2}{10xy - 2x - 56y + 12} \\ y(-3) &= 0 \end{align} This can be written as an exact equation. First we find the general solution following the paradigm.

1. We write the equation in the standard form, M dx + N dy = 0. $$(5y^2 + 6x - 2y + 2) dx + (10xy - 2x - 56y + 12) dy = 0$$
2. We test for exactness. $$\frac{\partial}{\partial y}\left(5y^2 + 6x - 2y + 2\right) = 10y - 2 = \frac{\partial}{\partial x}\left(10xy - 2x - 56y + 12\right)$$ so the equation is exact.

3. Write the partial differential equations. \begin{align} \frac{\partial F}{\partial x} &= 5y^2 + 6x - 2y + 2\\ \frac{\partial F}{\partial y} &= 10xy - 2x - 56y + 12 \end{align}
4. Integrate the first partial differential equation. $$F(x,y) = \int (5y^2 + 6x - 2y + 2)\,\partial x = 5xy^2 + 3x^2 - 2xy + 2x + C(y)$$
5. Integrate the second partial differential equation. $$F(x,y) = \int (10xy - 2x - 56y + 12)\,\partial y = 5xy^2 - 2xy - 28y^2 + 12y + \tilde{C}(x)$$
6. Equate the expressions for F(x,y).

Matching the expressions up, we find $C(y) = -28y^2 + 12y$ and $\tilde{C}(x) = 3x^2 + 2x.$ So $$F(x,y) = 5xy^2 - 28y^2 - 2xy + 3x^2 + 2x + 12y.$$

7. The solution is $F(x,y) = K.$ $$5xy^2 - 28y^2 - 2xy + 3x^2 + 2x + 12y = K$$
Now we plug in the initial values $x = -3$ and $y = 0$ and solve for $K = 21$. So the solution to the initial value problem is $$5xy^2 - 28y^2 - 2xy + 3x^2 + 2x + 12y = 21$$ You may reload this page to generate additional examples.