Kansas State University

Mathematics Department

Textbook Contents

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{y^2 - 2y - 6}{-9y^2 - 2xy - 16y + 2x + 15} \\ y(1) &= -3 \end{align}$$ 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. $$ (-y^2 + 2y + 6) dx + (-9y^2 - 2xy - 16y + 2x + 15) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(-y^2 + 2y + 6\right) = -2y + 2 = \frac{\partial}{\partial x}\left(-9y^2 - 2xy - 16y + 2x + 15\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= -y^2 + 2y + 6\\ \frac{\partial F}{\partial y} &= -9y^2 - 2xy - 16y + 2x + 15 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (-y^2 + 2y + 6)\,\partial x = -xy^2 + 2xy + 6x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (-9y^2 - 2xy - 16y + 2x + 15)\,\partial y = -3y^3 - xy^2 - 8y^2 + 2xy + 15y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = -3y^3 - 8y^2 + 15y$ and $ \tilde{C}(x) = 6x. $ So $$ F(x,y) = -xy^2 - 3y^3 + 2xy - 8y^2 + 6x + 15y. $$

  7. The solution is $F(x,y) = K.$ $$ -xy^2 - 3y^3 + 2xy - 8y^2 + 6x + 15y = K $$
Now we plug in the initial values $x = 1$ and $y = -3$ and solve for $K = -45$. So the solution to the initial value problem is $$ -xy^2 - 3y^3 + 2xy - 8y^2 + 6x + 15y = -45 $$ You may reload this page to generate additional examples.

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