Kansas State University

Mathematics Department

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{50xy + 34x - 10y - 10}{9y^2 - 25x^2 + 2y + 10x + 9} \\ y(0) &= 1 \end{align}$$ 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. $$ (-50xy - 34x + 10y + 10) dx + (9y^2 - 25x^2 + 2y + 10x + 9) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(-50xy - 34x + 10y + 10\right) = -50x + 10 = \frac{\partial}{\partial x}\left(9y^2 - 25x^2 + 2y + 10x + 9\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= -50xy - 34x + 10y + 10\\ \frac{\partial F}{\partial y} &= 9y^2 - 25x^2 + 2y + 10x + 9 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (-50xy - 34x + 10y + 10)\,\partial x = -25x^2y - 17x^2 + 10xy + 10x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (9y^2 - 25x^2 + 2y + 10x + 9)\,\partial y = 3y^3 - 25x^2y + y^2 + 10xy + 9y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = 3y^3 + y^2 + 9y$ and $ \tilde{C}(x) = -17x^2 + 10x. $ So $$ F(x,y) = 3y^3 - 25x^2y + 10xy - 17x^2 + y^2 + 10x + 9y. $$

  7. The solution is $F(x,y) = K.$ $$ 3y^3 - 25x^2y + 10xy - 17x^2 + y^2 + 10x + 9y = K $$
Now we plug in the initial values $x = 0$ and $y = 1$ and solve for $K = 13$. So the solution to the initial value problem is $$ 3y^3 - 25x^2y + 10xy - 17x^2 + y^2 + 10x + 9y = 13 $$ You may reload this page to generate additional examples.

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