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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{4y - 12}{3y^2 - 4x - 2y + 9} \\ y(1) &= -2 \end{align}$$ 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. $$ (-4y + 12) dx + (3y^2 - 4x - 2y + 9) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(-4y + 12\right) = -4 = \frac{\partial}{\partial x}\left(3y^2 - 4x - 2y + 9\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= -4y + 12\\ \frac{\partial F}{\partial y} &= 3y^2 - 4x - 2y + 9 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (-4y + 12)\,\partial x = -4xy + 12x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (3y^2 - 4x - 2y + 9)\,\partial y = y^3 - 4xy - y^2 + 9y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = y^3 - y^2 + 9y$ and $ \tilde{C}(x) = 12x. $ So $$ F(x,y) = y^3 - 4xy - y^2 + 9y + 12x. $$

  7. The solution is $F(x,y) = K.$ $$ y^3 - 4xy - y^2 + 9y + 12x = K $$
Now we plug in the initial values $x = 1$ and $y = -2$ and solve for $K = -10$. So the solution to the initial value problem is $$ y^3 - 4xy - y^2 + 9y + 12x = -10 $$ You may reload this page to generate additional examples.


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