Kansas State University

Mathematics Department

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{2x + 4y - 2}{-4x - 5} \\ y(2) &= -4 \end{align}$$ 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. $$ (-2x - 4y + 2) dx + (-4x - 5) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(-2x - 4y + 2\right) = -4 = \frac{\partial}{\partial x}\left(-4x - 5\right) $$ so the equation is exact.

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

    Matching the expressions up, we find $C(y) = -5y$ and $ \tilde{C}(x) = -x^2 + 2x. $ So $$ F(x,y) = -x^2 - 4xy + 2x - 5y. $$

  7. The solution is $F(x,y) = K.$ $$ -x^2 - 4xy + 2x - 5y = K $$
Now we plug in the initial values $x = 2$ and $y = -4$ and solve for $K = 52$. So the solution to the initial value problem is $$ -x^2 - 4xy + 2x - 5y = 52 $$ You may reload this page to generate additional examples.

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