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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{-8x + 20y + 10}{9y^2 - 20x - 8y + 30} \\ y(0) &= 2 \end{align}$$ 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. $$ (8x - 20y - 10) dx + (9y^2 - 20x - 8y + 30) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(8x - 20y - 10\right) = -20 = \frac{\partial}{\partial x}\left(9y^2 - 20x - 8y + 30\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= 8x - 20y - 10\\ \frac{\partial F}{\partial y} &= 9y^2 - 20x - 8y + 30 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (8x - 20y - 10)\,\partial x = 4x^2 - 20xy - 10x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (9y^2 - 20x - 8y + 30)\,\partial y = 3y^3 - 20xy - 4y^2 + 30y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = 3y^3 - 4y^2 + 30y$ and $ \tilde{C}(x) = 4x^2 - 10x. $ So $$ F(x,y) = 3y^3 - 4y^2 + 4x^2 - 20xy - 10x + 30y. $$

  7. The solution is $F(x,y) = K.$ $$ 3y^3 - 4y^2 + 4x^2 - 20xy - 10x + 30y = K $$
Now we plug in the initial values $x = 0$ and $y = 2$ and solve for $K = 68$. So the solution to the initial value problem is $$ 3y^3 - 4y^2 + 4x^2 - 20xy - 10x + 30y = 68 $$ You may reload this page to generate additional examples.


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