Kansas State University

Mathematics Department

Textbook Contents

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{9y^2 - 9y + 4x + 7}{-6y^2 - 18xy + 18y + 9x - 9} \\ y(2) &= 1 \end{align}$$ 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. $$ (-9y^2 + 9y - 4x - 7) dx + (-6y^2 - 18xy + 18y + 9x - 9) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(-9y^2 + 9y - 4x - 7\right) = -18y + 9 = \frac{\partial}{\partial x}\left(-6y^2 - 18xy + 18y + 9x - 9\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= -9y^2 + 9y - 4x - 7\\ \frac{\partial F}{\partial y} &= -6y^2 - 18xy + 18y + 9x - 9 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (-9y^2 + 9y - 4x - 7)\,\partial x = -9xy^2 + 9xy - 2x^2 - 7x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (-6y^2 - 18xy + 18y + 9x - 9)\,\partial y = -2y^3 - 9xy^2 + 9y^2 + 9xy - 9y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = -2y^3 + 9y^2 - 9y$ and $ \tilde{C}(x) = -2x^2 - 7x. $ So $$ F(x,y) = -2y^3 - 9xy^2 + 9xy - 2x^2 + 9y^2 - 7x - 9y. $$

  7. The solution is $F(x,y) = K.$ $$ -2y^3 - 9xy^2 + 9xy - 2x^2 + 9y^2 - 7x - 9y = K $$
Now we plug in the initial values $x = 2$ and $y = 1$ and solve for $K = -24$. So the solution to the initial value problem is $$ -2y^3 - 9xy^2 + 9xy - 2x^2 + 9y^2 - 7x - 9y = -24 $$ You may reload this page to generate additional examples.

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