Kansas State University

Mathematics Department

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{16xy - 24x - 10y + 23}{-6y^2 - 8x^2 - 8y + 10x - 7} \\ y(1) &= -2 \end{align}$$ 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. $$ (-16xy + 24x + 10y - 23) dx + (-6y^2 - 8x^2 - 8y + 10x - 7) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(-16xy + 24x + 10y - 23\right) = -16x + 10 = \frac{\partial}{\partial x}\left(-6y^2 - 8x^2 - 8y + 10x - 7\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= -16xy + 24x + 10y - 23\\ \frac{\partial F}{\partial y} &= -6y^2 - 8x^2 - 8y + 10x - 7 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (-16xy + 24x + 10y - 23)\,\partial x = -8x^2y + 12x^2 + 10xy - 23x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (-6y^2 - 8x^2 - 8y + 10x - 7)\,\partial y = -2y^3 - 8x^2y - 4y^2 + 10xy - 7y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = -2y^3 - 4y^2 - 7y$ and $ \tilde{C}(x) = 12x^2 - 23x. $ So $$ F(x,y) = -2y^3 - 8x^2y + 10xy + 12x^2 - 4y^2 - 23x - 7y. $$

  7. The solution is $F(x,y) = K.$ $$ -2y^3 - 8x^2y + 10xy + 12x^2 - 4y^2 - 23x - 7y = K $$
Now we plug in the initial values $x = 1$ and $y = -2$ and solve for $K = -1$. So the solution to the initial value problem is $$ -2y^3 - 8x^2y + 10xy + 12x^2 - 4y^2 - 23x - 7y = -1 $$ You may reload this page to generate additional examples.

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