Kansas State University

Mathematics Department

Textbook Contents

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{2xy + 18x + y + 9}{-9y^2 - x^2 - 6y - x + 1} \\ y(-3) &= 2 \end{align}$$ 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. $$ (-2xy - 18x - y - 9) dx + (-9y^2 - x^2 - 6y - x + 1) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(-2xy - 18x - y - 9\right) = -2x - 1 = \frac{\partial}{\partial x}\left(-9y^2 - x^2 - 6y - x + 1\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= -2xy - 18x - y - 9\\ \frac{\partial F}{\partial y} &= -9y^2 - x^2 - 6y - x + 1 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (-2xy - 18x - y - 9)\,\partial x = -x^2y - 9x^2 - xy - 9x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (-9y^2 - x^2 - 6y - x + 1)\,\partial y = -3y^3 - x^2y - 3y^2 - xy + y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = -3y^3 - 3y^2 + y$ and $ \tilde{C}(x) = -9x^2 - 9x. $ So $$ F(x,y) = -3y^3 - x^2y - xy - 9x^2 - 3y^2 - 9x + y. $$

  7. The solution is $F(x,y) = K.$ $$ -3y^3 - x^2y - xy - 9x^2 - 3y^2 - 9x + y = K $$
Now we plug in the initial values $x = -3$ and $y = 2$ and solve for $K = -100$. So the solution to the initial value problem is $$ -3y^3 - x^2y - xy - 9x^2 - 3y^2 - 9x + y = -100 $$ You may reload this page to generate additional examples.

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