Kansas State University

Mathematics Department

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Exact Equations

Additional Examples

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

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

    Matching the expressions up, we find $C(y) = 13y$ and $ \tilde{C}(x) = -4x^2 + 6x. $ So $$ F(x,y) = -4x^2 + 10xy + 6x + 13y. $$

  7. The solution is $F(x,y) = K.$ $$ -4x^2 + 10xy + 6x + 13y = K $$
Now we plug in the initial values $x = 1$ and $y = -2$ and solve for $K = -44$. So the solution to the initial value problem is $$ -4x^2 + 10xy + 6x + 13y = -44 $$ You may reload this page to generate additional examples.

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