Kansas State University

Mathematics Department

Textbook Contents

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{10y^2 + 4y + 5}{-20xy - 4x + 12y + 9} \\ y(-1) &= -3 \end{align}$$ 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. $$ (-10y^2 - 4y - 5) dx + (-20xy - 4x + 12y + 9) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(-10y^2 - 4y - 5\right) = -20y - 4 = \frac{\partial}{\partial x}\left(-20xy - 4x + 12y + 9\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= -10y^2 - 4y - 5\\ \frac{\partial F}{\partial y} &= -20xy - 4x + 12y + 9 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (-10y^2 - 4y - 5)\,\partial x = -10xy^2 - 4xy - 5x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (-20xy - 4x + 12y + 9)\,\partial y = -10xy^2 - 4xy + 6y^2 + 9y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = 6y^2 + 9y$ and $ \tilde{C}(x) = -5x. $ So $$ F(x,y) = -10xy^2 - 4xy + 6y^2 - 5x + 9y. $$

  7. The solution is $F(x,y) = K.$ $$ -10xy^2 - 4xy + 6y^2 - 5x + 9y = K $$
Now we plug in the initial values $x = -1$ and $y = -3$ and solve for $K = 110$. So the solution to the initial value problem is $$ -10xy^2 - 4xy + 6y^2 - 5x + 9y = 110 $$ You may reload this page to generate additional examples.

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