Kansas State University

Mathematics Department

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{10xy - 12x - 5y + 4}{-9y^2 - 5x^2 + 6y + 5x - 12} \\ y(2) &= -1 \end{align}$$ 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. $$ (-10xy + 12x + 5y - 4) dx + (-9y^2 - 5x^2 + 6y + 5x - 12) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(-10xy + 12x + 5y - 4\right) = -10x + 5 = \frac{\partial}{\partial x}\left(-9y^2 - 5x^2 + 6y + 5x - 12\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= -10xy + 12x + 5y - 4\\ \frac{\partial F}{\partial y} &= -9y^2 - 5x^2 + 6y + 5x - 12 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (-10xy + 12x + 5y - 4)\,\partial x = -5x^2y + 6x^2 + 5xy - 4x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (-9y^2 - 5x^2 + 6y + 5x - 12)\,\partial y = -3y^3 - 5x^2y + 3y^2 + 5xy - 12y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = -3y^3 + 3y^2 - 12y$ and $ \tilde{C}(x) = 6x^2 - 4x. $ So $$ F(x,y) = -3y^3 - 5x^2y + 5xy + 6x^2 + 3y^2 - 4x - 12y. $$

  7. The solution is $F(x,y) = K.$ $$ -3y^3 - 5x^2y + 5xy + 6x^2 + 3y^2 - 4x - 12y = K $$
Now we plug in the initial values $x = 2$ and $y = -1$ and solve for $K = 44$. So the solution to the initial value problem is $$ -3y^3 - 5x^2y + 5xy + 6x^2 + 3y^2 - 4x - 12y = 44 $$ You may reload this page to generate additional examples.

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