Kansas State University

Mathematics Department

Textbook Contents

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{20y^2 + 25y + 29}{3y^2 - 40xy - 30y - 25x - 27} \\ y(-2) &= -1 \end{align}$$ 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. $$ (-20y^2 - 25y - 29) dx + (3y^2 - 40xy - 30y - 25x - 27) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(-20y^2 - 25y - 29\right) = -40y - 25 = \frac{\partial}{\partial x}\left(3y^2 - 40xy - 30y - 25x - 27\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= -20y^2 - 25y - 29\\ \frac{\partial F}{\partial y} &= 3y^2 - 40xy - 30y - 25x - 27 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (-20y^2 - 25y - 29)\,\partial x = -20xy^2 - 25xy - 29x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (3y^2 - 40xy - 30y - 25x - 27)\,\partial y = y^3 - 20xy^2 - 15y^2 - 25xy - 27y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = y^3 - 15y^2 - 27y$ and $ \tilde{C}(x) = -29x. $ So $$ F(x,y) = -20xy^2 + y^3 - 25xy - 15y^2 - 29x - 27y. $$

  7. The solution is $F(x,y) = K.$ $$ -20xy^2 + y^3 - 25xy - 15y^2 - 29x - 27y = K $$
Now we plug in the initial values $x = -2$ and $y = -1$ and solve for $K = 59$. So the solution to the initial value problem is $$ -20xy^2 + y^3 - 25xy - 15y^2 - 29x - 27y = 59 $$ You may reload this page to generate additional examples.

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