Kansas State University

Mathematics Department

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{-16xy - 6x - 8y - 9}{-3y^2 + 8x^2 + 4y + 8x} \\ y(-3) &= 1 \end{align}$$ 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. $$ (16xy + 6x + 8y + 9) dx + (-3y^2 + 8x^2 + 4y + 8x) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(16xy + 6x + 8y + 9\right) = 16x + 8 = \frac{\partial}{\partial x}\left(-3y^2 + 8x^2 + 4y + 8x\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= 16xy + 6x + 8y + 9\\ \frac{\partial F}{\partial y} &= -3y^2 + 8x^2 + 4y + 8x \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (16xy + 6x + 8y + 9)\,\partial x = 8x^2y + 3x^2 + 8xy + 9x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (-3y^2 + 8x^2 + 4y + 8x)\,\partial y = -y^3 + 8x^2y + 2y^2 + 8xy + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = -y^3 + 2y^2$ and $ \tilde{C}(x) = 3x^2 + 9x. $ So $$ F(x,y) = 8x^2y - y^3 + 8xy + 3x^2 + 2y^2 + 9x. $$

  7. The solution is $F(x,y) = K.$ $$ 8x^2y - y^3 + 8xy + 3x^2 + 2y^2 + 9x = K $$
Now we plug in the initial values $x = -3$ and $y = 1$ and solve for $K = 49$. So the solution to the initial value problem is $$ 8x^2y - y^3 + 8xy + 3x^2 + 2y^2 + 9x = 49 $$ You may reload this page to generate additional examples.

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