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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{-25y^2 - 5y - 24}{50xy - 12y^2 - 44y + 5x - 6} \\ y(2) &= -4 \end{align}$$ 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. $$ (25y^2 + 5y + 24) dx + (50xy - 12y^2 - 44y + 5x - 6) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(25y^2 + 5y + 24\right) = 50y + 5 = \frac{\partial}{\partial x}\left(50xy - 12y^2 - 44y + 5x - 6\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= 25y^2 + 5y + 24\\ \frac{\partial F}{\partial y} &= 50xy - 12y^2 - 44y + 5x - 6 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (25y^2 + 5y + 24)\,\partial x = 25xy^2 + 5xy + 24x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (50xy - 12y^2 - 44y + 5x - 6)\,\partial y = 25xy^2 - 4y^3 - 22y^2 + 5xy - 6y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = -4y^3 - 22y^2 - 6y$ and $ \tilde{C}(x) = 24x. $ So $$ F(x,y) = -4y^3 + 25xy^2 + 5xy - 22y^2 - 6y + 24x. $$

  7. The solution is $F(x,y) = K.$ $$ -4y^3 + 25xy^2 + 5xy - 22y^2 - 6y + 24x = K $$
Now we plug in the initial values $x = 2$ and $y = -4$ and solve for $K = 736$. So the solution to the initial value problem is $$ -4y^3 + 25xy^2 + 5xy - 22y^2 - 6y + 24x = 736 $$ You may reload this page to generate additional examples.


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