Kansas State University

Mathematics Department

Textbook Contents

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{-4y^2 - 2y - 2x - 1}{8xy + 2x - 8y + 1} \\ y(-3) &= 0 \end{align}$$ 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. $$ (4y^2 + 2y + 2x + 1) dx + (8xy + 2x - 8y + 1) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(4y^2 + 2y + 2x + 1\right) = 8y + 2 = \frac{\partial}{\partial x}\left(8xy + 2x - 8y + 1\right) $$ so the equation is exact.

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

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

  7. The solution is $F(x,y) = K.$ $$ 4xy^2 + 2xy + x^2 - 4y^2 + x + y = K $$
Now we plug in the initial values $x = -3$ and $y = 0$ and solve for $K = 6$. So the solution to the initial value problem is $$ 4xy^2 + 2xy + x^2 - 4y^2 + x + y = 6 $$ You may reload this page to generate additional examples.

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