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### Exact Equations

Solve the following initial value problem, \begin{align} \frac{dy}{dx} &= \frac{-30xy - 28x - 20y - 24}{15x^2 + 6y^2 + 4y + 20x - 6} \\ y(0) &= -3 \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. $$(30xy + 28x + 20y + 24) dx + (15x^2 + 6y^2 + 4y + 20x - 6) dy = 0$$
2. We test for exactness. $$\frac{\partial}{\partial y}\left(30xy + 28x + 20y + 24\right) = 30x + 20 = \frac{\partial}{\partial x}\left(15x^2 + 6y^2 + 4y + 20x - 6\right)$$ so the equation is exact.

3. Write the partial differential equations. \begin{align} \frac{\partial F}{\partial x} &= 30xy + 28x + 20y + 24\\ \frac{\partial F}{\partial y} &= 15x^2 + 6y^2 + 4y + 20x - 6 \end{align}
4. Integrate the first partial differential equation. $$F(x,y) = \int (30xy + 28x + 20y + 24)\,\partial x = 15x^2y + 14x^2 + 20xy + 24x + C(y)$$
5. Integrate the second partial differential equation. $$F(x,y) = \int (15x^2 + 6y^2 + 4y + 20x - 6)\,\partial y = 15x^2y + 2y^3 + 2y^2 + 20xy - 6y + \tilde{C}(x)$$
6. Equate the expressions for F(x,y).

Matching the expressions up, we find $C(y) = 2y^3 + 2y^2 - 6y$ and $\tilde{C}(x) = 14x^2 + 24x.$ So $$F(x,y) = 2y^3 + 15x^2y + 14x^2 + 20xy + 2y^2 - 6y + 24x.$$

7. The solution is $F(x,y) = K.$ $$2y^3 + 15x^2y + 14x^2 + 20xy + 2y^2 - 6y + 24x = K$$
Now we plug in the initial values $x = 0$ and $y = -3$ and solve for $K = -18$. So the solution to the initial value problem is $$2y^3 + 15x^2y + 14x^2 + 20xy + 2y^2 - 6y + 24x = -18$$ You may reload this page to generate additional examples.