Kansas State University

Mathematics Department

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{5y - 2}{-5x + 6y + 15} \\ y(-2) &= -2 \end{align}$$ 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. $$ (-5y + 2) dx + (-5x + 6y + 15) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(-5y + 2\right) = -5 = \frac{\partial}{\partial x}\left(-5x + 6y + 15\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= -5y + 2\\ \frac{\partial F}{\partial y} &= -5x + 6y + 15 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (-5y + 2)\,\partial x = -5xy + 2x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (-5x + 6y + 15)\,\partial y = -5xy + 3y^2 + 15y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = 3y^2 + 15y$ and $ \tilde{C}(x) = 2x. $ So $$ F(x,y) = -5xy + 3y^2 + 2x + 15y. $$

  7. The solution is $F(x,y) = K.$ $$ -5xy + 3y^2 + 2x + 15y = K $$
Now we plug in the initial values $x = -2$ and $y = -2$ and solve for $K = -42$. So the solution to the initial value problem is $$ -5xy + 3y^2 + 2x + 15y = -42 $$ You may reload this page to generate additional examples.

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