Kansas State University

Mathematics Department

Textbook Contents

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{6xy + 4x - 3y - 4}{-3x^2 + 3x + 10} \\ y(-3) &= -2 \end{align}$$ 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. $$ (-6xy - 4x + 3y + 4) dx + (-3x^2 + 3x + 10) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(-6xy - 4x + 3y + 4\right) = -6x + 3 = \frac{\partial}{\partial x}\left(-3x^2 + 3x + 10\right) $$ so the equation is exact.

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

    Matching the expressions up, we find $C(y) = 10y$ and $ \tilde{C}(x) = -2x^2 + 4x. $ So $$ F(x,y) = -3x^2y - 2x^2 + 3xy + 4x + 10y. $$

  7. The solution is $F(x,y) = K.$ $$ -3x^2y - 2x^2 + 3xy + 4x + 10y = K $$
Now we plug in the initial values $x = -3$ and $y = -2$ and solve for $K = 22$. So the solution to the initial value problem is $$ -3x^2y - 2x^2 + 3xy + 4x + 10y = 22 $$ You may reload this page to generate additional examples.

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