Kansas State University

Mathematics Department

Textbook Contents

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{-3y^2 + 9y + 2x + 14}{6xy - 9x + 8y - 11} \\ y(-3) &= -3 \end{align}$$ 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. $$ (3y^2 - 9y - 2x - 14) dx + (6xy - 9x + 8y - 11) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(3y^2 - 9y - 2x - 14\right) = 6y - 9 = \frac{\partial}{\partial x}\left(6xy - 9x + 8y - 11\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= 3y^2 - 9y - 2x - 14\\ \frac{\partial F}{\partial y} &= 6xy - 9x + 8y - 11 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (3y^2 - 9y - 2x - 14)\,\partial x = 3xy^2 - 9xy - x^2 - 14x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (6xy - 9x + 8y - 11)\,\partial y = 3xy^2 - 9xy + 4y^2 - 11y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = 4y^2 - 11y$ and $ \tilde{C}(x) = -x^2 - 14x. $ So $$ F(x,y) = 3xy^2 - 9xy - x^2 + 4y^2 - 14x - 11y. $$

  7. The solution is $F(x,y) = K.$ $$ 3xy^2 - 9xy - x^2 + 4y^2 - 14x - 11y = K $$
Now we plug in the initial values $x = -3$ and $y = -3$ and solve for $K = -60$. So the solution to the initial value problem is $$ 3xy^2 - 9xy - x^2 + 4y^2 - 14x - 11y = -60 $$ You may reload this page to generate additional examples.

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