Kansas State University

Mathematics Department

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{10x - 2y + 13}{2x + 10y + 6} \\ y(1) &= -4 \end{align}$$ 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. $$ (-10x + 2y - 13) dx + (2x + 10y + 6) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(-10x + 2y - 13\right) = 2 = \frac{\partial}{\partial x}\left(2x + 10y + 6\right) $$ so the equation is exact.

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

    Matching the expressions up, we find $C(y) = 5y^2 + 6y$ and $ \tilde{C}(x) = -5x^2 - 13x. $ So $$ F(x,y) = -5x^2 + 2xy + 5y^2 - 13x + 6y. $$

  7. The solution is $F(x,y) = K.$ $$ -5x^2 + 2xy + 5y^2 - 13x + 6y = K $$
Now we plug in the initial values $x = 1$ and $y = -4$ and solve for $K = 30$. So the solution to the initial value problem is $$ -5x^2 + 2xy + 5y^2 - 13x + 6y = 30 $$ You may reload this page to generate additional examples.

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