Kansas State University

Mathematics Department

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{-10y^2 + 4y + 10x - 5}{15y^2 + 20xy - 30y - 4x - 1} \\ y(1) &= 0 \end{align}$$ 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. $$ (10y^2 - 4y - 10x + 5) dx + (15y^2 + 20xy - 30y - 4x - 1) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(10y^2 - 4y - 10x + 5\right) = 20y - 4 = \frac{\partial}{\partial x}\left(15y^2 + 20xy - 30y - 4x - 1\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= 10y^2 - 4y - 10x + 5\\ \frac{\partial F}{\partial y} &= 15y^2 + 20xy - 30y - 4x - 1 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (10y^2 - 4y - 10x + 5)\,\partial x = 10xy^2 - 4xy - 5x^2 + 5x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (15y^2 + 20xy - 30y - 4x - 1)\,\partial y = 5y^3 + 10xy^2 - 15y^2 - 4xy - y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = 5y^3 - 15y^2 - y$ and $ \tilde{C}(x) = -5x^2 + 5x. $ So $$ F(x,y) = 5y^3 + 10xy^2 - 4xy - 5x^2 - 15y^2 + 5x - y. $$

  7. The solution is $F(x,y) = K.$ $$ 5y^3 + 10xy^2 - 4xy - 5x^2 - 15y^2 + 5x - y = K $$
Now we plug in the initial values $x = 1$ and $y = 0$ and solve for $K = 0$. So the solution to the initial value problem is $$ 5y^3 + 10xy^2 - 4xy - 5x^2 - 15y^2 + 5x - y = 0 $$ You may reload this page to generate additional examples.

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