Kansas State University

Mathematics Department

Textbook Contents

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{8y^2 + 20y - 2x + 25}{-16xy - 20x + 26y + 26} \\ y(0) &= 1 \end{align}$$ 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. $$ (-8y^2 - 20y + 2x - 25) dx + (-16xy - 20x + 26y + 26) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(-8y^2 - 20y + 2x - 25\right) = -16y - 20 = \frac{\partial}{\partial x}\left(-16xy - 20x + 26y + 26\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= -8y^2 - 20y + 2x - 25\\ \frac{\partial F}{\partial y} &= -16xy - 20x + 26y + 26 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (-8y^2 - 20y + 2x - 25)\,\partial x = -8xy^2 - 20xy + x^2 - 25x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (-16xy - 20x + 26y + 26)\,\partial y = -8xy^2 - 20xy + 13y^2 + 26y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = 13y^2 + 26y$ and $ \tilde{C}(x) = x^2 - 25x. $ So $$ F(x,y) = -8xy^2 - 20xy + x^2 + 13y^2 - 25x + 26y. $$

  7. The solution is $F(x,y) = K.$ $$ -8xy^2 - 20xy + x^2 + 13y^2 - 25x + 26y = K $$
Now we plug in the initial values $x = 0$ and $y = 1$ and solve for $K = 39$. So the solution to the initial value problem is $$ -8xy^2 - 20xy + x^2 + 13y^2 - 25x + 26y = 39 $$ You may reload this page to generate additional examples.

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