Kansas State University

Mathematics Department

Textbook Contents

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{6y^2 + 10y - 11}{-12xy - 10x - 40y - 23} \\ y(-1) &= -1 \end{align}$$ 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. $$ (-6y^2 - 10y + 11) dx + (-12xy - 10x - 40y - 23) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(-6y^2 - 10y + 11\right) = -12y - 10 = \frac{\partial}{\partial x}\left(-12xy - 10x - 40y - 23\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= -6y^2 - 10y + 11\\ \frac{\partial F}{\partial y} &= -12xy - 10x - 40y - 23 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (-6y^2 - 10y + 11)\,\partial x = -6xy^2 - 10xy + 11x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (-12xy - 10x - 40y - 23)\,\partial y = -6xy^2 - 10xy - 20y^2 - 23y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = -20y^2 - 23y$ and $ \tilde{C}(x) = 11x. $ So $$ F(x,y) = -6xy^2 - 10xy - 20y^2 + 11x - 23y. $$

  7. The solution is $F(x,y) = K.$ $$ -6xy^2 - 10xy - 20y^2 + 11x - 23y = K $$
Now we plug in the initial values $x = -1$ and $y = -1$ and solve for $K = -12$. So the solution to the initial value problem is $$ -6xy^2 - 10xy - 20y^2 + 11x - 23y = -12 $$ You may reload this page to generate additional examples.

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