Kansas State University

Mathematics Department

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{15y^2 - 9y - 2x - 8}{-30xy + 9x - 30y + 8} \\ y(0) &= 0 \end{align}$$ 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. $$ (-15y^2 + 9y + 2x + 8) dx + (-30xy + 9x - 30y + 8) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(-15y^2 + 9y + 2x + 8\right) = -30y + 9 = \frac{\partial}{\partial x}\left(-30xy + 9x - 30y + 8\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= -15y^2 + 9y + 2x + 8\\ \frac{\partial F}{\partial y} &= -30xy + 9x - 30y + 8 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (-15y^2 + 9y + 2x + 8)\,\partial x = -15xy^2 + 9xy + x^2 + 8x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (-30xy + 9x - 30y + 8)\,\partial y = -15xy^2 + 9xy - 15y^2 + 8y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = -15y^2 + 8y$ and $ \tilde{C}(x) = x^2 + 8x. $ So $$ F(x,y) = -15xy^2 + 9xy + x^2 - 15y^2 + 8x + 8y. $$

  7. The solution is $F(x,y) = K.$ $$ -15xy^2 + 9xy + x^2 - 15y^2 + 8x + 8y = K $$
Now we plug in the initial values $x = 0$ and $y = 0$ and solve for $K = 0$. So the solution to the initial value problem is $$ -15xy^2 + 9xy + x^2 - 15y^2 + 8x + 8y = 0 $$ You may reload this page to generate additional examples.

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