Kansas State University

Mathematics Department

Textbook Contents

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{8x + 6y + 18}{-6x + 6y + 6} \\ y(0) &= -2 \end{align}$$ 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. $$ (-8x - 6y - 18) dx + (-6x + 6y + 6) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(-8x - 6y - 18\right) = -6 = \frac{\partial}{\partial x}\left(-6x + 6y + 6\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= -8x - 6y - 18\\ \frac{\partial F}{\partial y} &= -6x + 6y + 6 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (-8x - 6y - 18)\,\partial x = -4x^2 - 6xy - 18x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (-6x + 6y + 6)\,\partial y = -6xy + 3y^2 + 6y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = 3y^2 + 6y$ and $ \tilde{C}(x) = -4x^2 - 18x. $ So $$ F(x,y) = -4x^2 - 6xy + 3y^2 - 18x + 6y. $$

  7. The solution is $F(x,y) = K.$ $$ -4x^2 - 6xy + 3y^2 - 18x + 6y = K $$
Now we plug in the initial values $x = 0$ and $y = -2$ and solve for $K = 0$. So the solution to the initial value problem is $$ -4x^2 - 6xy + 3y^2 - 18x + 6y = 0 $$ You may reload this page to generate additional examples.

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