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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{-30xy - 28x - 20y - 24}{15x^2 + 6y^2 + 4y + 20x - 6} \\ y(0) &= -3 \end{align}$$ 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. $$ (30xy + 28x + 20y + 24) dx + (15x^2 + 6y^2 + 4y + 20x - 6) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(30xy + 28x + 20y + 24\right) = 30x + 20 = \frac{\partial}{\partial x}\left(15x^2 + 6y^2 + 4y + 20x - 6\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= 30xy + 28x + 20y + 24\\ \frac{\partial F}{\partial y} &= 15x^2 + 6y^2 + 4y + 20x - 6 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (30xy + 28x + 20y + 24)\,\partial x = 15x^2y + 14x^2 + 20xy + 24x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (15x^2 + 6y^2 + 4y + 20x - 6)\,\partial y = 15x^2y + 2y^3 + 2y^2 + 20xy - 6y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = 2y^3 + 2y^2 - 6y$ and $ \tilde{C}(x) = 14x^2 + 24x. $ So $$ F(x,y) = 2y^3 + 15x^2y + 14x^2 + 20xy + 2y^2 - 6y + 24x. $$

  7. The solution is $F(x,y) = K.$ $$ 2y^3 + 15x^2y + 14x^2 + 20xy + 2y^2 - 6y + 24x = K $$
Now we plug in the initial values $x = 0$ and $y = -3$ and solve for $K = -18$. So the solution to the initial value problem is $$ 2y^3 + 15x^2y + 14x^2 + 20xy + 2y^2 - 6y + 24x = -18 $$ You may reload this page to generate additional examples.


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