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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{5y^2 - 10x + 2y + 10}{-10xy - 9y^2 + 38y - 2x + 1} \\ y(2) &= 4 \end{align}$$ 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. $$ (-5y^2 + 10x - 2y - 10) dx + (-10xy - 9y^2 + 38y - 2x + 1) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(-5y^2 + 10x - 2y - 10\right) = -10y - 2 = \frac{\partial}{\partial x}\left(-10xy - 9y^2 + 38y - 2x + 1\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= -5y^2 + 10x - 2y - 10\\ \frac{\partial F}{\partial y} &= -10xy - 9y^2 + 38y - 2x + 1 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (-5y^2 + 10x - 2y - 10)\,\partial x = -5xy^2 + 5x^2 - 2xy - 10x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (-10xy - 9y^2 + 38y - 2x + 1)\,\partial y = -5xy^2 - 3y^3 + 19y^2 - 2xy + y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = -3y^3 + 19y^2 + y$ and $ \tilde{C}(x) = 5x^2 - 10x. $ So $$ F(x,y) = -5xy^2 - 3y^3 + 19y^2 + 5x^2 - 2xy - 10x + y. $$

  7. The solution is $F(x,y) = K.$ $$ -5xy^2 - 3y^3 + 19y^2 + 5x^2 - 2xy - 10x + y = K $$
Now we plug in the initial values $x = 2$ and $y = 4$ and solve for $K = -60$. So the solution to the initial value problem is $$ -5xy^2 - 3y^3 + 19y^2 + 5x^2 - 2xy - 10x + y = -60 $$ You may reload this page to generate additional examples.


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