Kansas State University

Mathematics Department

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{25y^2 - 25y - 8x - 16}{-50xy + 25x - 52y + 20} \\ y(0) &= 3 \end{align}$$ 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. $$ (-25y^2 + 25y + 8x + 16) dx + (-50xy + 25x - 52y + 20) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(-25y^2 + 25y + 8x + 16\right) = -50y + 25 = \frac{\partial}{\partial x}\left(-50xy + 25x - 52y + 20\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= -25y^2 + 25y + 8x + 16\\ \frac{\partial F}{\partial y} &= -50xy + 25x - 52y + 20 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (-25y^2 + 25y + 8x + 16)\,\partial x = -25xy^2 + 25xy + 4x^2 + 16x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (-50xy + 25x - 52y + 20)\,\partial y = -25xy^2 + 25xy - 26y^2 + 20y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = -26y^2 + 20y$ and $ \tilde{C}(x) = 4x^2 + 16x. $ So $$ F(x,y) = -25xy^2 + 25xy + 4x^2 - 26y^2 + 16x + 20y. $$

  7. The solution is $F(x,y) = K.$ $$ -25xy^2 + 25xy + 4x^2 - 26y^2 + 16x + 20y = K $$
Now we plug in the initial values $x = 0$ and $y = 3$ and solve for $K = -174$. So the solution to the initial value problem is $$ -25xy^2 + 25xy + 4x^2 - 26y^2 + 16x + 20y = -174 $$ You may reload this page to generate additional examples.

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