Kansas State University

Mathematics Department

Textbook Contents

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{8y^2 - 20y + 2x + 16}{-16xy + 20x - 24y + 22} \\ y(-3) &= 4 \end{align}$$ 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. $$ (-8y^2 + 20y - 2x - 16) dx + (-16xy + 20x - 24y + 22) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(-8y^2 + 20y - 2x - 16\right) = -16y + 20 = \frac{\partial}{\partial x}\left(-16xy + 20x - 24y + 22\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= -8y^2 + 20y - 2x - 16\\ \frac{\partial F}{\partial y} &= -16xy + 20x - 24y + 22 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (-8y^2 + 20y - 2x - 16)\,\partial x = -8xy^2 + 20xy - x^2 - 16x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (-16xy + 20x - 24y + 22)\,\partial y = -8xy^2 + 20xy - 12y^2 + 22y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = -12y^2 + 22y$ and $ \tilde{C}(x) = -x^2 - 16x. $ So $$ F(x,y) = -8xy^2 + 20xy - x^2 - 12y^2 - 16x + 22y. $$

  7. The solution is $F(x,y) = K.$ $$ -8xy^2 + 20xy - x^2 - 12y^2 - 16x + 22y = K $$
Now we plug in the initial values $x = -3$ and $y = 4$ and solve for $K = 79$. So the solution to the initial value problem is $$ -8xy^2 + 20xy - x^2 - 12y^2 - 16x + 22y = 79 $$ You may reload this page to generate additional examples.

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