Kansas State University

Mathematics Department

Textbook Contents

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{-16xy - 4x - 16y - 7}{8x^2 + 16x + 13} \\ y(-3) &= 1 \end{align}$$ 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. $$ (16xy + 4x + 16y + 7) dx + (8x^2 + 16x + 13) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(16xy + 4x + 16y + 7\right) = 16x + 16 = \frac{\partial}{\partial x}\left(8x^2 + 16x + 13\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= 16xy + 4x + 16y + 7\\ \frac{\partial F}{\partial y} &= 8x^2 + 16x + 13 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (16xy + 4x + 16y + 7)\,\partial x = 8x^2y + 2x^2 + 16xy + 7x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (8x^2 + 16x + 13)\,\partial y = 8x^2y + 16xy + 13y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = 13y$ and $ \tilde{C}(x) = 2x^2 + 7x. $ So $$ F(x,y) = 8x^2y + 2x^2 + 16xy + 7x + 13y. $$

  7. The solution is $F(x,y) = K.$ $$ 8x^2y + 2x^2 + 16xy + 7x + 13y = K $$
Now we plug in the initial values $x = -3$ and $y = 1$ and solve for $K = 34$. So the solution to the initial value problem is $$ 8x^2y + 2x^2 + 16xy + 7x + 13y = 34 $$ You may reload this page to generate additional examples.

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