Kansas State University

Mathematics Department

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{-50xy - 40x + 5y + 3}{25x^2 - 5x - 10y + 4} \\ y(2) &= -2 \end{align}$$ 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. $$ (50xy + 40x - 5y - 3) dx + (25x^2 - 5x - 10y + 4) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(50xy + 40x - 5y - 3\right) = 50x - 5 = \frac{\partial}{\partial x}\left(25x^2 - 5x - 10y + 4\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= 50xy + 40x - 5y - 3\\ \frac{\partial F}{\partial y} &= 25x^2 - 5x - 10y + 4 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (50xy + 40x - 5y - 3)\,\partial x = 25x^2y + 20x^2 - 5xy - 3x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (25x^2 - 5x - 10y + 4)\,\partial y = 25x^2y - 5xy - 5y^2 + 4y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = -5y^2 + 4y$ and $ \tilde{C}(x) = 20x^2 - 3x. $ So $$ F(x,y) = 25x^2y - 5xy + 20x^2 - 5y^2 - 3x + 4y. $$

  7. The solution is $F(x,y) = K.$ $$ 25x^2y - 5xy + 20x^2 - 5y^2 - 3x + 4y = K $$
Now we plug in the initial values $x = 2$ and $y = -2$ and solve for $K = -134$. So the solution to the initial value problem is $$ 25x^2y - 5xy + 20x^2 - 5y^2 - 3x + 4y = -134 $$ You may reload this page to generate additional examples.

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