Kansas State University

Mathematics Department

Textbook Contents

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{-2xy + 2x - 2y - 6}{-9y^2 + x^2 - 6y + 2x} \\ y(-3) &= -2 \end{align}$$ 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. $$ (2xy - 2x + 2y + 6) dx + (-9y^2 + x^2 - 6y + 2x) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(2xy - 2x + 2y + 6\right) = 2x + 2 = \frac{\partial}{\partial x}\left(-9y^2 + x^2 - 6y + 2x\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= 2xy - 2x + 2y + 6\\ \frac{\partial F}{\partial y} &= -9y^2 + x^2 - 6y + 2x \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (2xy - 2x + 2y + 6)\,\partial x = x^2y - x^2 + 2xy + 6x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (-9y^2 + x^2 - 6y + 2x)\,\partial y = -3y^3 + x^2y - 3y^2 + 2xy + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

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

  7. The solution is $F(x,y) = K.$ $$ x^2y - 3y^3 + 2xy - x^2 - 3y^2 + 6x = K $$
Now we plug in the initial values $x = -3$ and $y = -2$ and solve for $K = -21$. So the solution to the initial value problem is $$ x^2y - 3y^3 + 2xy - x^2 - 3y^2 + 6x = -21 $$ You may reload this page to generate additional examples.

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