Kansas State University

Mathematics Department

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{40xy - 22x - 15y + 9}{-20x^2 + 15x - 10y + 10} \\ y(-3) &= -1 \end{align}$$ 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. $$ (-40xy + 22x + 15y - 9) dx + (-20x^2 + 15x - 10y + 10) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(-40xy + 22x + 15y - 9\right) = -40x + 15 = \frac{\partial}{\partial x}\left(-20x^2 + 15x - 10y + 10\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= -40xy + 22x + 15y - 9\\ \frac{\partial F}{\partial y} &= -20x^2 + 15x - 10y + 10 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (-40xy + 22x + 15y - 9)\,\partial x = -20x^2y + 11x^2 + 15xy - 9x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (-20x^2 + 15x - 10y + 10)\,\partial y = -20x^2y + 15xy - 5y^2 + 10y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = -5y^2 + 10y$ and $ \tilde{C}(x) = 11x^2 - 9x. $ So $$ F(x,y) = -20x^2y + 15xy + 11x^2 - 5y^2 - 9x + 10y. $$

  7. The solution is $F(x,y) = K.$ $$ -20x^2y + 15xy + 11x^2 - 5y^2 - 9x + 10y = K $$
Now we plug in the initial values $x = -3$ and $y = -1$ and solve for $K = 336$. So the solution to the initial value problem is $$ -20x^2y + 15xy + 11x^2 - 5y^2 - 9x + 10y = 336 $$ You may reload this page to generate additional examples.

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