Kansas State University

Mathematics Department

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{-8y^2 + 20y - 10x - 2}{-15y^2 + 16xy - 12y - 20x + 23} \\ y(1) &= -3 \end{align}$$ 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. $$ (8y^2 - 20y + 10x + 2) dx + (-15y^2 + 16xy - 12y - 20x + 23) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(8y^2 - 20y + 10x + 2\right) = 16y - 20 = \frac{\partial}{\partial x}\left(-15y^2 + 16xy - 12y - 20x + 23\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= 8y^2 - 20y + 10x + 2\\ \frac{\partial F}{\partial y} &= -15y^2 + 16xy - 12y - 20x + 23 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (8y^2 - 20y + 10x + 2)\,\partial x = 8xy^2 - 20xy + 5x^2 + 2x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (-15y^2 + 16xy - 12y - 20x + 23)\,\partial y = -5y^3 + 8xy^2 - 6y^2 - 20xy + 23y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = -5y^3 - 6y^2 + 23y$ and $ \tilde{C}(x) = 5x^2 + 2x. $ So $$ F(x,y) = -5y^3 + 8xy^2 - 20xy + 5x^2 - 6y^2 + 2x + 23y. $$

  7. The solution is $F(x,y) = K.$ $$ -5y^3 + 8xy^2 - 20xy + 5x^2 - 6y^2 + 2x + 23y = K $$
Now we plug in the initial values $x = 1$ and $y = -3$ and solve for $K = 151$. So the solution to the initial value problem is $$ -5y^3 + 8xy^2 - 20xy + 5x^2 - 6y^2 + 2x + 23y = 151 $$ You may reload this page to generate additional examples.

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