Kansas State University

Mathematics Department

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{-2xy + 2x + 3y - 8}{9y^2 + x^2 - 4y - 3x - 5} \\ y(-2) &= -1 \end{align}$$ 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. $$ (2xy - 2x - 3y + 8) dx + (9y^2 + x^2 - 4y - 3x - 5) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(2xy - 2x - 3y + 8\right) = 2x - 3 = \frac{\partial}{\partial x}\left(9y^2 + x^2 - 4y - 3x - 5\right) $$ so the equation is exact.

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

    Matching the expressions up, we find $C(y) = 3y^3 - 2y^2 - 5y$ and $ \tilde{C}(x) = -x^2 + 8x. $ So $$ F(x,y) = 3y^3 + x^2y - 3xy - x^2 - 2y^2 + 8x - 5y. $$

  7. The solution is $F(x,y) = K.$ $$ 3y^3 + x^2y - 3xy - x^2 - 2y^2 + 8x - 5y = K $$
Now we plug in the initial values $x = -2$ and $y = -1$ and solve for $K = -30$. So the solution to the initial value problem is $$ 3y^3 + x^2y - 3xy - x^2 - 2y^2 + 8x - 5y = -30 $$ You may reload this page to generate additional examples.

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