Kansas State University

Mathematics Department

Textbook Contents

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{-25y^2 + 15y + 5}{50xy - 15x - 24y + 11} \\ y(0) &= -1 \end{align}$$ 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. $$ (25y^2 - 15y - 5) dx + (50xy - 15x - 24y + 11) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(25y^2 - 15y - 5\right) = 50y - 15 = \frac{\partial}{\partial x}\left(50xy - 15x - 24y + 11\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= 25y^2 - 15y - 5\\ \frac{\partial F}{\partial y} &= 50xy - 15x - 24y + 11 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (25y^2 - 15y - 5)\,\partial x = 25xy^2 - 15xy - 5x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (50xy - 15x - 24y + 11)\,\partial y = 25xy^2 - 15xy - 12y^2 + 11y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = -12y^2 + 11y$ and $ \tilde{C}(x) = -5x. $ So $$ F(x,y) = 25xy^2 - 15xy - 12y^2 - 5x + 11y. $$

  7. The solution is $F(x,y) = K.$ $$ 25xy^2 - 15xy - 12y^2 - 5x + 11y = K $$
Now we plug in the initial values $x = 0$ and $y = -1$ and solve for $K = -23$. So the solution to the initial value problem is $$ 25xy^2 - 15xy - 12y^2 - 5x + 11y = -23 $$ You may reload this page to generate additional examples.

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