Kansas State University

Mathematics Department

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{40xy - 48x - 5y + 6}{-20x^2 + 5x - 8y - 17} \\ y(2) &= 0 \end{align}$$ 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. $$ (-40xy + 48x + 5y - 6) dx + (-20x^2 + 5x - 8y - 17) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(-40xy + 48x + 5y - 6\right) = -40x + 5 = \frac{\partial}{\partial x}\left(-20x^2 + 5x - 8y - 17\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= -40xy + 48x + 5y - 6\\ \frac{\partial F}{\partial y} &= -20x^2 + 5x - 8y - 17 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (-40xy + 48x + 5y - 6)\,\partial x = -20x^2y + 24x^2 + 5xy - 6x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (-20x^2 + 5x - 8y - 17)\,\partial y = -20x^2y + 5xy - 4y^2 - 17y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = -4y^2 - 17y$ and $ \tilde{C}(x) = 24x^2 - 6x. $ So $$ F(x,y) = -20x^2y + 5xy + 24x^2 - 4y^2 - 6x - 17y. $$

  7. The solution is $F(x,y) = K.$ $$ -20x^2y + 5xy + 24x^2 - 4y^2 - 6x - 17y = K $$
Now we plug in the initial values $x = 2$ and $y = 0$ and solve for $K = 84$. So the solution to the initial value problem is $$ -20x^2y + 5xy + 24x^2 - 4y^2 - 6x - 17y = 84 $$ You may reload this page to generate additional examples.

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