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Exact Equations

Solve the following initial value problem, \begin{align} \frac{dy}{dx} &= \frac{-5y^2 - 10x + 3y}{10xy - 12y^2 - 34y - 3x + 14} \\ y(-2) &= 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. $$(5y^2 + 10x - 3y) dx + (10xy - 12y^2 - 34y - 3x + 14) dy = 0$$
2. We test for exactness. $$\frac{\partial}{\partial y}\left(5y^2 + 10x - 3y\right) = 10y - 3 = \frac{\partial}{\partial x}\left(10xy - 12y^2 - 34y - 3x + 14\right)$$ so the equation is exact.

3. Write the partial differential equations. \begin{align} \frac{\partial F}{\partial x} &= 5y^2 + 10x - 3y\\ \frac{\partial F}{\partial y} &= 10xy - 12y^2 - 34y - 3x + 14 \end{align}
4. Integrate the first partial differential equation. $$F(x,y) = \int (5y^2 + 10x - 3y)\,\partial x = 5xy^2 + 5x^2 - 3xy + C(y)$$
5. Integrate the second partial differential equation. $$F(x,y) = \int (10xy - 12y^2 - 34y - 3x + 14)\,\partial y = 5xy^2 - 4y^3 - 17y^2 - 3xy + 14y + \tilde{C}(x)$$
6. Equate the expressions for F(x,y).

Matching the expressions up, we find $C(y) = -4y^3 - 17y^2 + 14y$ and $\tilde{C}(x) = 5x^2.$ So $$F(x,y) = -4y^3 + 5xy^2 - 17y^2 + 5x^2 - 3xy + 14y.$$

7. The solution is $F(x,y) = K.$ $$-4y^3 + 5xy^2 - 17y^2 + 5x^2 - 3xy + 14y = K$$
Now we plug in the initial values $x = -2$ and $y = 0$ and solve for $K = 20$. So the solution to the initial value problem is $$-4y^3 + 5xy^2 - 17y^2 + 5x^2 - 3xy + 14y = 20$$ You may reload this page to generate additional examples.