Kansas State University

Mathematics Department

Textbook Contents

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{24xy + 44x + 9y + 12}{-12x^2 - 9x + 4} \\ y(-3) &= -3 \end{align}$$ 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. $$ (-24xy - 44x - 9y - 12) dx + (-12x^2 - 9x + 4) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(-24xy - 44x - 9y - 12\right) = -24x - 9 = \frac{\partial}{\partial x}\left(-12x^2 - 9x + 4\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= -24xy - 44x - 9y - 12\\ \frac{\partial F}{\partial y} &= -12x^2 - 9x + 4 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (-24xy - 44x - 9y - 12)\,\partial x = -12x^2y - 22x^2 - 9xy - 12x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (-12x^2 - 9x + 4)\,\partial y = -12x^2y - 9xy + 4y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = 4y$ and $ \tilde{C}(x) = -22x^2 - 12x. $ So $$ F(x,y) = -12x^2y - 22x^2 - 9xy - 12x + 4y. $$

  7. The solution is $F(x,y) = K.$ $$ -12x^2y - 22x^2 - 9xy - 12x + 4y = K $$
Now we plug in the initial values $x = -3$ and $y = -3$ and solve for $K = 69$. So the solution to the initial value problem is $$ -12x^2y - 22x^2 - 9xy - 12x + 4y = 69 $$ You may reload this page to generate additional examples.

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