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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{-2x + 4y + 4}{12y^2 - 4x - 10y + 5} \\ y(-1) &= 1 \end{align}$$ 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. $$ (2x - 4y - 4) dx + (12y^2 - 4x - 10y + 5) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(2x - 4y - 4\right) = -4 = \frac{\partial}{\partial x}\left(12y^2 - 4x - 10y + 5\right) $$ so the equation is exact.

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

    Matching the expressions up, we find $C(y) = 4y^3 - 5y^2 + 5y$ and $ \tilde{C}(x) = x^2 - 4x. $ So $$ F(x,y) = 4y^3 - 5y^2 + x^2 - 4xy - 4x + 5y. $$

  7. The solution is $F(x,y) = K.$ $$ 4y^3 - 5y^2 + x^2 - 4xy - 4x + 5y = K $$
Now we plug in the initial values $x = -1$ and $y = 1$ and solve for $K = 13$. So the solution to the initial value problem is $$ 4y^3 - 5y^2 + x^2 - 4xy - 4x + 5y = 13 $$ You may reload this page to generate additional examples.


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