Kansas State University

Mathematics Department

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{2xy - 6x - 4y + 12}{-6y^2 - x^2 - 10y + 4x + 10} \\ y(-2) &= -4 \end{align}$$ 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. $$ (-2xy + 6x + 4y - 12) dx + (-6y^2 - x^2 - 10y + 4x + 10) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(-2xy + 6x + 4y - 12\right) = -2x + 4 = \frac{\partial}{\partial x}\left(-6y^2 - x^2 - 10y + 4x + 10\right) $$ so the equation is exact.

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

    Matching the expressions up, we find $C(y) = -2y^3 - 5y^2 + 10y$ and $ \tilde{C}(x) = 3x^2 - 12x. $ So $$ F(x,y) = -2y^3 - x^2y + 4xy + 3x^2 - 5y^2 - 12x + 10y. $$

  7. The solution is $F(x,y) = K.$ $$ -2y^3 - x^2y + 4xy + 3x^2 - 5y^2 - 12x + 10y = K $$
Now we plug in the initial values $x = -2$ and $y = -4$ and solve for $K = 92$. So the solution to the initial value problem is $$ -2y^3 - x^2y + 4xy + 3x^2 - 5y^2 - 12x + 10y = 92 $$ You may reload this page to generate additional examples.

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©2010, 2014 Andrew G. Bennett