Kansas State University

Mathematics Department

Textbook Contents

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{-3y^2 + 2y + 6x + 5}{6xy - 2x - 26y + 12} \\ y(1) &= 1 \end{align}$$ 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. $$ (3y^2 - 2y - 6x - 5) dx + (6xy - 2x - 26y + 12) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(3y^2 - 2y - 6x - 5\right) = 6y - 2 = \frac{\partial}{\partial x}\left(6xy - 2x - 26y + 12\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= 3y^2 - 2y - 6x - 5\\ \frac{\partial F}{\partial y} &= 6xy - 2x - 26y + 12 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (3y^2 - 2y - 6x - 5)\,\partial x = 3xy^2 - 2xy - 3x^2 - 5x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (6xy - 2x - 26y + 12)\,\partial y = 3xy^2 - 2xy - 13y^2 + 12y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = -13y^2 + 12y$ and $ \tilde{C}(x) = -3x^2 - 5x. $ So $$ F(x,y) = 3xy^2 - 2xy - 3x^2 - 13y^2 - 5x + 12y. $$

  7. The solution is $F(x,y) = K.$ $$ 3xy^2 - 2xy - 3x^2 - 13y^2 - 5x + 12y = K $$
Now we plug in the initial values $x = 1$ and $y = 1$ and solve for $K = -8$. So the solution to the initial value problem is $$ 3xy^2 - 2xy - 3x^2 - 13y^2 - 5x + 12y = -8 $$ You may reload this page to generate additional examples.

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