Kansas State University

Mathematics Department

Textbook Contents

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{-30xy + 10x + 9y - 4}{15x^2 - 9x + 6y + 11} \\ y(0) &= 4 \end{align}$$ 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. $$ (30xy - 10x - 9y + 4) dx + (15x^2 - 9x + 6y + 11) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(30xy - 10x - 9y + 4\right) = 30x - 9 = \frac{\partial}{\partial x}\left(15x^2 - 9x + 6y + 11\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= 30xy - 10x - 9y + 4\\ \frac{\partial F}{\partial y} &= 15x^2 - 9x + 6y + 11 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (30xy - 10x - 9y + 4)\,\partial x = 15x^2y - 5x^2 - 9xy + 4x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (15x^2 - 9x + 6y + 11)\,\partial y = 15x^2y - 9xy + 3y^2 + 11y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = 3y^2 + 11y$ and $ \tilde{C}(x) = -5x^2 + 4x. $ So $$ F(x,y) = 15x^2y - 9xy - 5x^2 + 3y^2 + 4x + 11y. $$

  7. The solution is $F(x,y) = K.$ $$ 15x^2y - 9xy - 5x^2 + 3y^2 + 4x + 11y = K $$
Now we plug in the initial values $x = 0$ and $y = 4$ and solve for $K = 92$. So the solution to the initial value problem is $$ 15x^2y - 9xy - 5x^2 + 3y^2 + 4x + 11y = 92 $$ You may reload this page to generate additional examples.

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