Kansas State University

Mathematics Department

Textbook Contents

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{-15y^2 - 25y - 29}{30xy + 25x - 30y - 22} \\ y(0) &= -2 \end{align}$$ 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. $$ (15y^2 + 25y + 29) dx + (30xy + 25x - 30y - 22) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(15y^2 + 25y + 29\right) = 30y + 25 = \frac{\partial}{\partial x}\left(30xy + 25x - 30y - 22\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= 15y^2 + 25y + 29\\ \frac{\partial F}{\partial y} &= 30xy + 25x - 30y - 22 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (15y^2 + 25y + 29)\,\partial x = 15xy^2 + 25xy + 29x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (30xy + 25x - 30y - 22)\,\partial y = 15xy^2 + 25xy - 15y^2 - 22y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = -15y^2 - 22y$ and $ \tilde{C}(x) = 29x. $ So $$ F(x,y) = 15xy^2 + 25xy - 15y^2 + 29x - 22y. $$

  7. The solution is $F(x,y) = K.$ $$ 15xy^2 + 25xy - 15y^2 + 29x - 22y = K $$
Now we plug in the initial values $x = 0$ and $y = -2$ and solve for $K = -16$. So the solution to the initial value problem is $$ 15xy^2 + 25xy - 15y^2 + 29x - 22y = -16 $$ You may reload this page to generate additional examples.

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