Kansas State University

Mathematics Department

Textbook Contents

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{-2y^2 + 2y + 4x - 3}{6y^2 + 4xy + 14y - 2x - 6} \\ y(-2) &= -3 \end{align}$$ 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. $$ (2y^2 - 2y - 4x + 3) dx + (6y^2 + 4xy + 14y - 2x - 6) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(2y^2 - 2y - 4x + 3\right) = 4y - 2 = \frac{\partial}{\partial x}\left(6y^2 + 4xy + 14y - 2x - 6\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= 2y^2 - 2y - 4x + 3\\ \frac{\partial F}{\partial y} &= 6y^2 + 4xy + 14y - 2x - 6 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (2y^2 - 2y - 4x + 3)\,\partial x = 2xy^2 - 2xy - 2x^2 + 3x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (6y^2 + 4xy + 14y - 2x - 6)\,\partial y = 2y^3 + 2xy^2 + 7y^2 - 2xy - 6y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = 2y^3 + 7y^2 - 6y$ and $ \tilde{C}(x) = -2x^2 + 3x. $ So $$ F(x,y) = 2y^3 + 2xy^2 - 2xy - 2x^2 + 7y^2 + 3x - 6y. $$

  7. The solution is $F(x,y) = K.$ $$ 2y^3 + 2xy^2 - 2xy - 2x^2 + 7y^2 + 3x - 6y = K $$
Now we plug in the initial values $x = -2$ and $y = -3$ and solve for $K = -35$. So the solution to the initial value problem is $$ 2y^3 + 2xy^2 - 2xy - 2x^2 + 7y^2 + 3x - 6y = -35 $$ You may reload this page to generate additional examples.

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