Kansas State University

Mathematics Department

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{16xy - 16y + 8x}{-8x^2 + 16x - 6y + 19} \\ y(0) &= -2 \end{align}$$ 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. $$ (-16xy + 16y - 8x) dx + (-8x^2 + 16x - 6y + 19) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(-16xy + 16y - 8x\right) = -16x + 16 = \frac{\partial}{\partial x}\left(-8x^2 + 16x - 6y + 19\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= -16xy + 16y - 8x\\ \frac{\partial F}{\partial y} &= -8x^2 + 16x - 6y + 19 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (-16xy + 16y - 8x)\,\partial x = -8x^2y + 16xy - 4x^2 + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (-8x^2 + 16x - 6y + 19)\,\partial y = -8x^2y + 16xy - 3y^2 + 19y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = -3y^2 + 19y$ and $ \tilde{C}(x) = -4x^2. $ So $$ F(x,y) = -8x^2y + 16xy - 4x^2 - 3y^2 + 19y. $$

  7. The solution is $F(x,y) = K.$ $$ -8x^2y + 16xy - 4x^2 - 3y^2 + 19y = K $$
Now we plug in the initial values $x = 0$ and $y = -2$ and solve for $K = -50$. So the solution to the initial value problem is $$ -8x^2y + 16xy - 4x^2 - 3y^2 + 19y = -50 $$ You may reload this page to generate additional examples.

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