Kansas State University

Mathematics Department

Textbook Contents

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{8xy - 14x + 8y - 7}{-4x^2 - 8x - 10y - 21} \\ y(-2) &= 0 \end{align}$$ 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. $$ (-8xy + 14x - 8y + 7) dx + (-4x^2 - 8x - 10y - 21) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(-8xy + 14x - 8y + 7\right) = -8x - 8 = \frac{\partial}{\partial x}\left(-4x^2 - 8x - 10y - 21\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= -8xy + 14x - 8y + 7\\ \frac{\partial F}{\partial y} &= -4x^2 - 8x - 10y - 21 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (-8xy + 14x - 8y + 7)\,\partial x = -4x^2y + 7x^2 - 8xy + 7x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (-4x^2 - 8x - 10y - 21)\,\partial y = -4x^2y - 8xy - 5y^2 - 21y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = -5y^2 - 21y$ and $ \tilde{C}(x) = 7x^2 + 7x. $ So $$ F(x,y) = -4x^2y - 8xy + 7x^2 - 5y^2 + 7x - 21y. $$

  7. The solution is $F(x,y) = K.$ $$ -4x^2y - 8xy + 7x^2 - 5y^2 + 7x - 21y = K $$
Now we plug in the initial values $x = -2$ and $y = 0$ and solve for $K = 14$. So the solution to the initial value problem is $$ -4x^2y - 8xy + 7x^2 - 5y^2 + 7x - 21y = 14 $$ You may reload this page to generate additional examples.

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