Kansas State University

Mathematics Department

Textbook Contents

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{16y^2 + 12y - 2x - 6}{-32xy - 12x + 16y + 8} \\ y(1) &= 1 \end{align}$$ 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. $$ (-16y^2 - 12y + 2x + 6) dx + (-32xy - 12x + 16y + 8) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(-16y^2 - 12y + 2x + 6\right) = -32y - 12 = \frac{\partial}{\partial x}\left(-32xy - 12x + 16y + 8\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= -16y^2 - 12y + 2x + 6\\ \frac{\partial F}{\partial y} &= -32xy - 12x + 16y + 8 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (-16y^2 - 12y + 2x + 6)\,\partial x = -16xy^2 - 12xy + x^2 + 6x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (-32xy - 12x + 16y + 8)\,\partial y = -16xy^2 - 12xy + 8y^2 + 8y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = 8y^2 + 8y$ and $ \tilde{C}(x) = x^2 + 6x. $ So $$ F(x,y) = -16xy^2 - 12xy + x^2 + 8y^2 + 6x + 8y. $$

  7. The solution is $F(x,y) = K.$ $$ -16xy^2 - 12xy + x^2 + 8y^2 + 6x + 8y = K $$
Now we plug in the initial values $x = 1$ and $y = 1$ and solve for $K = -5$. So the solution to the initial value problem is $$ -16xy^2 - 12xy + x^2 + 8y^2 + 6x + 8y = -5 $$ You may reload this page to generate additional examples.

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