Kansas State University

Mathematics Department

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{-8y^2 + 8y + 10x - 1}{16xy - 8x + 40y - 17} \\ y(-1) &= -2 \end{align}$$ 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. $$ (8y^2 - 8y - 10x + 1) dx + (16xy - 8x + 40y - 17) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(8y^2 - 8y - 10x + 1\right) = 16y - 8 = \frac{\partial}{\partial x}\left(16xy - 8x + 40y - 17\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= 8y^2 - 8y - 10x + 1\\ \frac{\partial F}{\partial y} &= 16xy - 8x + 40y - 17 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (8y^2 - 8y - 10x + 1)\,\partial x = 8xy^2 - 8xy - 5x^2 + x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (16xy - 8x + 40y - 17)\,\partial y = 8xy^2 - 8xy + 20y^2 - 17y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = 20y^2 - 17y$ and $ \tilde{C}(x) = -5x^2 + x. $ So $$ F(x,y) = 8xy^2 - 8xy - 5x^2 + 20y^2 + x - 17y. $$

  7. The solution is $F(x,y) = K.$ $$ 8xy^2 - 8xy - 5x^2 + 20y^2 + x - 17y = K $$
Now we plug in the initial values $x = -1$ and $y = -2$ and solve for $K = 60$. So the solution to the initial value problem is $$ 8xy^2 - 8xy - 5x^2 + 20y^2 + x - 17y = 60 $$ You may reload this page to generate additional examples.

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