Kansas State University

Mathematics Department

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{-40xy + 30x + 8y - 8}{20x^2 - 8x + 2y - 13} \\ y(2) &= 0 \end{align}$$ 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. $$ (40xy - 30x - 8y + 8) dx + (20x^2 - 8x + 2y - 13) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(40xy - 30x - 8y + 8\right) = 40x - 8 = \frac{\partial}{\partial x}\left(20x^2 - 8x + 2y - 13\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= 40xy - 30x - 8y + 8\\ \frac{\partial F}{\partial y} &= 20x^2 - 8x + 2y - 13 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (40xy - 30x - 8y + 8)\,\partial x = 20x^2y - 15x^2 - 8xy + 8x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (20x^2 - 8x + 2y - 13)\,\partial y = 20x^2y - 8xy + y^2 - 13y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = y^2 - 13y$ and $ \tilde{C}(x) = -15x^2 + 8x. $ So $$ F(x,y) = 20x^2y - 8xy - 15x^2 + y^2 + 8x - 13y. $$

  7. The solution is $F(x,y) = K.$ $$ 20x^2y - 8xy - 15x^2 + y^2 + 8x - 13y = K $$
Now we plug in the initial values $x = 2$ and $y = 0$ and solve for $K = -44$. So the solution to the initial value problem is $$ 20x^2y - 8xy - 15x^2 + y^2 + 8x - 13y = -44 $$ You may reload this page to generate additional examples.

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