Kansas State University

Mathematics Department

Textbook Contents

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{-12xy + 20x + 2y - 5}{6x^2 - 2x - 4y + 7} \\ y(-1) &= 1 \end{align}$$ 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. $$ (12xy - 20x - 2y + 5) dx + (6x^2 - 2x - 4y + 7) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(12xy - 20x - 2y + 5\right) = 12x - 2 = \frac{\partial}{\partial x}\left(6x^2 - 2x - 4y + 7\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= 12xy - 20x - 2y + 5\\ \frac{\partial F}{\partial y} &= 6x^2 - 2x - 4y + 7 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (12xy - 20x - 2y + 5)\,\partial x = 6x^2y - 10x^2 - 2xy + 5x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (6x^2 - 2x - 4y + 7)\,\partial y = 6x^2y - 2xy - 2y^2 + 7y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = -2y^2 + 7y$ and $ \tilde{C}(x) = -10x^2 + 5x. $ So $$ F(x,y) = 6x^2y - 2xy - 10x^2 - 2y^2 + 5x + 7y. $$

  7. The solution is $F(x,y) = K.$ $$ 6x^2y - 2xy - 10x^2 - 2y^2 + 5x + 7y = K $$
Now we plug in the initial values $x = -1$ and $y = 1$ and solve for $K = -2$. So the solution to the initial value problem is $$ 6x^2y - 2xy - 10x^2 - 2y^2 + 5x + 7y = -2 $$ You may reload this page to generate additional examples.

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