Kansas State University

Mathematics Department

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{30xy - 40x - 12y + 20}{-6y^2 - 15x^2 - 2y + 12x + 2} \\ y(-2) &= 4 \end{align}$$ 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. $$ (-30xy + 40x + 12y - 20) dx + (-6y^2 - 15x^2 - 2y + 12x + 2) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(-30xy + 40x + 12y - 20\right) = -30x + 12 = \frac{\partial}{\partial x}\left(-6y^2 - 15x^2 - 2y + 12x + 2\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= -30xy + 40x + 12y - 20\\ \frac{\partial F}{\partial y} &= -6y^2 - 15x^2 - 2y + 12x + 2 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (-30xy + 40x + 12y - 20)\,\partial x = -15x^2y + 20x^2 + 12xy - 20x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (-6y^2 - 15x^2 - 2y + 12x + 2)\,\partial y = -2y^3 - 15x^2y - y^2 + 12xy + 2y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = -2y^3 - y^2 + 2y$ and $ \tilde{C}(x) = 20x^2 - 20x. $ So $$ F(x,y) = -2y^3 - 15x^2y + 12xy + 20x^2 - y^2 - 20x + 2y. $$

  7. The solution is $F(x,y) = K.$ $$ -2y^3 - 15x^2y + 12xy + 20x^2 - y^2 - 20x + 2y = K $$
Now we plug in the initial values $x = -2$ and $y = 4$ and solve for $K = -352$. So the solution to the initial value problem is $$ -2y^3 - 15x^2y + 12xy + 20x^2 - y^2 - 20x + 2y = -352 $$ You may reload this page to generate additional examples.

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