Kansas State University

Mathematics Department

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{4x + 10y + 6}{-12y^2 - 4y - 10x - 10} \\ y(-3) &= 1 \end{align}$$ 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. $$ (-4x - 10y - 6) dx + (-12y^2 - 4y - 10x - 10) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(-4x - 10y - 6\right) = -10 = \frac{\partial}{\partial x}\left(-12y^2 - 4y - 10x - 10\right) $$ so the equation is exact.

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

    Matching the expressions up, we find $C(y) = -4y^3 - 2y^2 - 10y$ and $ \tilde{C}(x) = -2x^2 - 6x. $ So $$ F(x,y) = -4y^3 - 10xy - 2x^2 - 2y^2 - 6x - 10y. $$

  7. The solution is $F(x,y) = K.$ $$ -4y^3 - 10xy - 2x^2 - 2y^2 - 6x - 10y = K $$
Now we plug in the initial values $x = -3$ and $y = 1$ and solve for $K = 14$. So the solution to the initial value problem is $$ -4y^3 - 10xy - 2x^2 - 2y^2 - 6x - 10y = 14 $$ You may reload this page to generate additional examples.

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