Kansas State University

Mathematics Department

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{-y^2 + 2y - 4x - 7}{2xy - 2x - 2y + 9} \\ y(-2) &= 2 \end{align}$$ 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. $$ (y^2 - 2y + 4x + 7) dx + (2xy - 2x - 2y + 9) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(y^2 - 2y + 4x + 7\right) = 2y - 2 = \frac{\partial}{\partial x}\left(2xy - 2x - 2y + 9\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= y^2 - 2y + 4x + 7\\ \frac{\partial F}{\partial y} &= 2xy - 2x - 2y + 9 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (y^2 - 2y + 4x + 7)\,\partial x = xy^2 - 2xy + 2x^2 + 7x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (2xy - 2x - 2y + 9)\,\partial y = xy^2 - 2xy - y^2 + 9y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = -y^2 + 9y$ and $ \tilde{C}(x) = 2x^2 + 7x. $ So $$ F(x,y) = xy^2 - 2xy + 2x^2 - y^2 + 7x + 9y. $$

  7. The solution is $F(x,y) = K.$ $$ xy^2 - 2xy + 2x^2 - y^2 + 7x + 9y = K $$
Now we plug in the initial values $x = -2$ and $y = 2$ and solve for $K = 8$. So the solution to the initial value problem is $$ xy^2 - 2xy + 2x^2 - y^2 + 7x + 9y = 8 $$ You may reload this page to generate additional examples.

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