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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{-10y^2 + 6x - 2y - 13}{20xy + 2x - 42y - 8} \\ y(1) &= 2 \end{align}$$ 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. $$ (10y^2 - 6x + 2y + 13) dx + (20xy + 2x - 42y - 8) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(10y^2 - 6x + 2y + 13\right) = 20y + 2 = \frac{\partial}{\partial x}\left(20xy + 2x - 42y - 8\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= 10y^2 - 6x + 2y + 13\\ \frac{\partial F}{\partial y} &= 20xy + 2x - 42y - 8 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (10y^2 - 6x + 2y + 13)\,\partial x = 10xy^2 - 3x^2 + 2xy + 13x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (20xy + 2x - 42y - 8)\,\partial y = 10xy^2 + 2xy - 21y^2 - 8y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = -21y^2 - 8y$ and $ \tilde{C}(x) = -3x^2 + 13x. $ So $$ F(x,y) = 10xy^2 - 21y^2 + 2xy - 3x^2 + 13x - 8y. $$

  7. The solution is $F(x,y) = K.$ $$ 10xy^2 - 21y^2 + 2xy - 3x^2 + 13x - 8y = K $$
Now we plug in the initial values $x = 1$ and $y = 2$ and solve for $K = -46$. So the solution to the initial value problem is $$ 10xy^2 - 21y^2 + 2xy - 3x^2 + 13x - 8y = -46 $$ You may reload this page to generate additional examples.


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