Kansas State University

Mathematics Department

Textbook Contents

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{16xy + 20x + 2y - 1}{-8x^2 - 2x + 2y + 3} \\ y(-1) &= -2 \end{align}$$ 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. $$ (-16xy - 20x - 2y + 1) dx + (-8x^2 - 2x + 2y + 3) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(-16xy - 20x - 2y + 1\right) = -16x - 2 = \frac{\partial}{\partial x}\left(-8x^2 - 2x + 2y + 3\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= -16xy - 20x - 2y + 1\\ \frac{\partial F}{\partial y} &= -8x^2 - 2x + 2y + 3 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (-16xy - 20x - 2y + 1)\,\partial x = -8x^2y - 10x^2 - 2xy + x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (-8x^2 - 2x + 2y + 3)\,\partial y = -8x^2y - 2xy + y^2 + 3y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = y^2 + 3y$ and $ \tilde{C}(x) = -10x^2 + x. $ So $$ F(x,y) = -8x^2y - 2xy - 10x^2 + y^2 + x + 3y. $$

  7. The solution is $F(x,y) = K.$ $$ -8x^2y - 2xy - 10x^2 + y^2 + x + 3y = K $$
Now we plug in the initial values $x = -1$ and $y = -2$ and solve for $K = -1$. So the solution to the initial value problem is $$ -8x^2y - 2xy - 10x^2 + y^2 + x + 3y = -1 $$ You may reload this page to generate additional examples.

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