Kansas State University

Mathematics Department

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{-8x + 16y + 22}{-16x - 4} \\ y(-3) &= 0 \end{align}$$ 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. $$ (8x - 16y - 22) dx + (-16x - 4) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(8x - 16y - 22\right) = -16 = \frac{\partial}{\partial x}\left(-16x - 4\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= 8x - 16y - 22\\ \frac{\partial F}{\partial y} &= -16x - 4 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (8x - 16y - 22)\,\partial x = 4x^2 - 16xy - 22x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (-16x - 4)\,\partial y = -16xy - 4y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = -4y$ and $ \tilde{C}(x) = 4x^2 - 22x. $ So $$ F(x,y) = 4x^2 - 16xy - 22x - 4y. $$

  7. The solution is $F(x,y) = K.$ $$ 4x^2 - 16xy - 22x - 4y = K $$
Now we plug in the initial values $x = -3$ and $y = 0$ and solve for $K = 102$. So the solution to the initial value problem is $$ 4x^2 - 16xy - 22x - 4y = 102 $$ You may reload this page to generate additional examples.

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