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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{-20y^2 - 6x - 4y - 1}{40xy + 4x + 30y + 9} \\ y(-1) &= 0 \end{align}$$ 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. $$ (20y^2 + 6x + 4y + 1) dx + (40xy + 4x + 30y + 9) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(20y^2 + 6x + 4y + 1\right) = 40y + 4 = \frac{\partial}{\partial x}\left(40xy + 4x + 30y + 9\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= 20y^2 + 6x + 4y + 1\\ \frac{\partial F}{\partial y} &= 40xy + 4x + 30y + 9 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (20y^2 + 6x + 4y + 1)\,\partial x = 20xy^2 + 3x^2 + 4xy + x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (40xy + 4x + 30y + 9)\,\partial y = 20xy^2 + 4xy + 15y^2 + 9y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = 15y^2 + 9y$ and $ \tilde{C}(x) = 3x^2 + x. $ So $$ F(x,y) = 20xy^2 + 15y^2 + 4xy + 3x^2 + x + 9y. $$

  7. The solution is $F(x,y) = K.$ $$ 20xy^2 + 15y^2 + 4xy + 3x^2 + x + 9y = K $$
Now we plug in the initial values $x = -1$ and $y = 0$ and solve for $K = 2$. So the solution to the initial value problem is $$ 20xy^2 + 15y^2 + 4xy + 3x^2 + x + 9y = 2 $$ You may reload this page to generate additional examples.


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