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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{12y^2 + 10x - 4y - 10}{-24xy + 4x - 6y - 4} \\ y(2) &= -4 \end{align}$$ 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. $$ (-12y^2 - 10x + 4y + 10) dx + (-24xy + 4x - 6y - 4) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(-12y^2 - 10x + 4y + 10\right) = -24y + 4 = \frac{\partial}{\partial x}\left(-24xy + 4x - 6y - 4\right) $$ so the equation is exact.

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

    Matching the expressions up, we find $C(y) = -3y^2 - 4y$ and $ \tilde{C}(x) = -5x^2 + 10x. $ So $$ F(x,y) = -12xy^2 - 3y^2 + 4xy - 5x^2 + 10x - 4y. $$

  7. The solution is $F(x,y) = K.$ $$ -12xy^2 - 3y^2 + 4xy - 5x^2 + 10x - 4y = K $$
Now we plug in the initial values $x = 2$ and $y = -4$ and solve for $K = -448$. So the solution to the initial value problem is $$ -12xy^2 - 3y^2 + 4xy - 5x^2 + 10x - 4y = -448 $$ You may reload this page to generate additional examples.


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