Kansas State University

Mathematics Department

Textbook Contents

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{2x - 12y - 14}{3y^2 - 8y + 12x + 11} \\ y(-3) &= 1 \end{align}$$ 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. $$ (-2x + 12y + 14) dx + (3y^2 - 8y + 12x + 11) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(-2x + 12y + 14\right) = 12 = \frac{\partial}{\partial x}\left(3y^2 - 8y + 12x + 11\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= -2x + 12y + 14\\ \frac{\partial F}{\partial y} &= 3y^2 - 8y + 12x + 11 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (-2x + 12y + 14)\,\partial x = -x^2 + 12xy + 14x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (3y^2 - 8y + 12x + 11)\,\partial y = y^3 - 4y^2 + 12xy + 11y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = y^3 - 4y^2 + 11y$ and $ \tilde{C}(x) = -x^2 + 14x. $ So $$ F(x,y) = y^3 + 12xy - x^2 - 4y^2 + 14x + 11y. $$

  7. The solution is $F(x,y) = K.$ $$ y^3 + 12xy - x^2 - 4y^2 + 14x + 11y = K $$
Now we plug in the initial values $x = -3$ and $y = 1$ and solve for $K = -79$. So the solution to the initial value problem is $$ y^3 + 12xy - x^2 - 4y^2 + 14x + 11y = -79 $$ You may reload this page to generate additional examples.

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