Kansas State University

Mathematics Department

Textbook Contents

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{-24xy + 40x + 6y - 8}{12x^2 - 6x + 10y - 12} \\ y(0) &= 1 \end{align}$$ 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. $$ (24xy - 40x - 6y + 8) dx + (12x^2 - 6x + 10y - 12) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(24xy - 40x - 6y + 8\right) = 24x - 6 = \frac{\partial}{\partial x}\left(12x^2 - 6x + 10y - 12\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= 24xy - 40x - 6y + 8\\ \frac{\partial F}{\partial y} &= 12x^2 - 6x + 10y - 12 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (24xy - 40x - 6y + 8)\,\partial x = 12x^2y - 20x^2 - 6xy + 8x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (12x^2 - 6x + 10y - 12)\,\partial y = 12x^2y - 6xy + 5y^2 - 12y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = 5y^2 - 12y$ and $ \tilde{C}(x) = -20x^2 + 8x. $ So $$ F(x,y) = 12x^2y - 6xy - 20x^2 + 5y^2 + 8x - 12y. $$

  7. The solution is $F(x,y) = K.$ $$ 12x^2y - 6xy - 20x^2 + 5y^2 + 8x - 12y = K $$
Now we plug in the initial values $x = 0$ and $y = 1$ and solve for $K = -7$. So the solution to the initial value problem is $$ 12x^2y - 6xy - 20x^2 + 5y^2 + 8x - 12y = -7 $$ You may reload this page to generate additional examples.

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