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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{-12y^2 - 16y - 13}{24xy + 6y^2 + 20y + 16x + 20} \\ y(1) &= -2 \end{align}$$ 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. $$ (12y^2 + 16y + 13) dx + (24xy + 6y^2 + 20y + 16x + 20) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(12y^2 + 16y + 13\right) = 24y + 16 = \frac{\partial}{\partial x}\left(24xy + 6y^2 + 20y + 16x + 20\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= 12y^2 + 16y + 13\\ \frac{\partial F}{\partial y} &= 24xy + 6y^2 + 20y + 16x + 20 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (12y^2 + 16y + 13)\,\partial x = 12xy^2 + 16xy + 13x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (24xy + 6y^2 + 20y + 16x + 20)\,\partial y = 12xy^2 + 2y^3 + 10y^2 + 16xy + 20y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = 2y^3 + 10y^2 + 20y$ and $ \tilde{C}(x) = 13x. $ So $$ F(x,y) = 2y^3 + 12xy^2 + 16xy + 10y^2 + 20y + 13x. $$

  7. The solution is $F(x,y) = K.$ $$ 2y^3 + 12xy^2 + 16xy + 10y^2 + 20y + 13x = K $$
Now we plug in the initial values $x = 1$ and $y = -2$ and solve for $K = 13$. So the solution to the initial value problem is $$ 2y^3 + 12xy^2 + 16xy + 10y^2 + 20y + 13x = 13 $$ You may reload this page to generate additional examples.


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