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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{10x - 4y - 3}{4x + 16} \\ y(2) &= -2 \end{align}$$ 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. $$ (-10x + 4y + 3) dx + (4x + 16) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(-10x + 4y + 3\right) = 4 = \frac{\partial}{\partial x}\left(4x + 16\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= -10x + 4y + 3\\ \frac{\partial F}{\partial y} &= 4x + 16 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (-10x + 4y + 3)\,\partial x = -5x^2 + 4xy + 3x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (4x + 16)\,\partial y = 4xy + 16y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = 16y$ and $ \tilde{C}(x) = -5x^2 + 3x. $ So $$ F(x,y) = -5x^2 + 4xy + 16y + 3x. $$

  7. The solution is $F(x,y) = K.$ $$ -5x^2 + 4xy + 16y + 3x = K $$
Now we plug in the initial values $x = 2$ and $y = -2$ and solve for $K = -62$. So the solution to the initial value problem is $$ -5x^2 + 4xy + 16y + 3x = -62 $$ You may reload this page to generate additional examples.


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