Kansas State University

Mathematics Department

Textbook Contents

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{-30xy - 20x + 5y - 3}{15x^2 - 5x + 10y - 9} \\ y(-3) &= 1 \end{align}$$ 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. $$ (30xy + 20x - 5y + 3) dx + (15x^2 - 5x + 10y - 9) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(30xy + 20x - 5y + 3\right) = 30x - 5 = \frac{\partial}{\partial x}\left(15x^2 - 5x + 10y - 9\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= 30xy + 20x - 5y + 3\\ \frac{\partial F}{\partial y} &= 15x^2 - 5x + 10y - 9 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (30xy + 20x - 5y + 3)\,\partial x = 15x^2y + 10x^2 - 5xy + 3x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (15x^2 - 5x + 10y - 9)\,\partial y = 15x^2y - 5xy + 5y^2 - 9y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = 5y^2 - 9y$ and $ \tilde{C}(x) = 10x^2 + 3x. $ So $$ F(x,y) = 15x^2y - 5xy + 10x^2 + 5y^2 + 3x - 9y. $$

  7. The solution is $F(x,y) = K.$ $$ 15x^2y - 5xy + 10x^2 + 5y^2 + 3x - 9y = K $$
Now we plug in the initial values $x = -3$ and $y = 1$ and solve for $K = 227$. So the solution to the initial value problem is $$ 15x^2y - 5xy + 10x^2 + 5y^2 + 3x - 9y = 227 $$ You may reload this page to generate additional examples.

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