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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{-6xy + 2x + 12y - 3}{3x^2 + 6y^2 + 6y - 12x - 13} \\ y(-1) &= -4 \end{align}$$ 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. $$ (6xy - 2x - 12y + 3) dx + (3x^2 + 6y^2 + 6y - 12x - 13) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(6xy - 2x - 12y + 3\right) = 6x - 12 = \frac{\partial}{\partial x}\left(3x^2 + 6y^2 + 6y - 12x - 13\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= 6xy - 2x - 12y + 3\\ \frac{\partial F}{\partial y} &= 3x^2 + 6y^2 + 6y - 12x - 13 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (6xy - 2x - 12y + 3)\,\partial x = 3x^2y - x^2 - 12xy + 3x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (3x^2 + 6y^2 + 6y - 12x - 13)\,\partial y = 3x^2y + 2y^3 + 3y^2 - 12xy - 13y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = 2y^3 + 3y^2 - 13y$ and $ \tilde{C}(x) = -x^2 + 3x. $ So $$ F(x,y) = 2y^3 + 3x^2y - x^2 - 12xy + 3y^2 - 13y + 3x. $$

  7. The solution is $F(x,y) = K.$ $$ 2y^3 + 3x^2y - x^2 - 12xy + 3y^2 - 13y + 3x = K $$
Now we plug in the initial values $x = -1$ and $y = -4$ and solve for $K = -92$. So the solution to the initial value problem is $$ 2y^3 + 3x^2y - x^2 - 12xy + 3y^2 - 13y + 3x = -92 $$ You may reload this page to generate additional examples.


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