Kansas State University

Mathematics Department

Textbook Contents

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{-8xy + 10x + 20y - 8}{4x^2 - 20x + 4y + 6} \\ y(-1) &= 2 \end{align}$$ 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. $$ (8xy - 10x - 20y + 8) dx + (4x^2 - 20x + 4y + 6) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(8xy - 10x - 20y + 8\right) = 8x - 20 = \frac{\partial}{\partial x}\left(4x^2 - 20x + 4y + 6\right) $$ so the equation is exact.

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

    Matching the expressions up, we find $C(y) = 2y^2 + 6y$ and $ \tilde{C}(x) = -5x^2 + 8x. $ So $$ F(x,y) = 4x^2y - 20xy - 5x^2 + 2y^2 + 8x + 6y. $$

  7. The solution is $F(x,y) = K.$ $$ 4x^2y - 20xy - 5x^2 + 2y^2 + 8x + 6y = K $$
Now we plug in the initial values $x = -1$ and $y = 2$ and solve for $K = 55$. So the solution to the initial value problem is $$ 4x^2y - 20xy - 5x^2 + 2y^2 + 8x + 6y = 55 $$ You may reload this page to generate additional examples.

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