Kansas State University

Mathematics Department

Textbook Contents

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{25y^2 + 5y + 10x + 22}{-50xy - 5x - 40y - 9} \\ y(1) &= 3 \end{align}$$ 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. $$ (-25y^2 - 5y - 10x - 22) dx + (-50xy - 5x - 40y - 9) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(-25y^2 - 5y - 10x - 22\right) = -50y - 5 = \frac{\partial}{\partial x}\left(-50xy - 5x - 40y - 9\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= -25y^2 - 5y - 10x - 22\\ \frac{\partial F}{\partial y} &= -50xy - 5x - 40y - 9 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (-25y^2 - 5y - 10x - 22)\,\partial x = -25xy^2 - 5xy - 5x^2 - 22x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (-50xy - 5x - 40y - 9)\,\partial y = -25xy^2 - 5xy - 20y^2 - 9y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = -20y^2 - 9y$ and $ \tilde{C}(x) = -5x^2 - 22x. $ So $$ F(x,y) = -25xy^2 - 5xy - 5x^2 - 20y^2 - 22x - 9y. $$

  7. The solution is $F(x,y) = K.$ $$ -25xy^2 - 5xy - 5x^2 - 20y^2 - 22x - 9y = K $$
Now we plug in the initial values $x = 1$ and $y = 3$ and solve for $K = -474$. So the solution to the initial value problem is $$ -25xy^2 - 5xy - 5x^2 - 20y^2 - 22x - 9y = -474 $$ You may reload this page to generate additional examples.

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