Kansas State University

Mathematics Department

Textbook Contents

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{30xy + 16x + 10y + 13}{-15x^2 - 10x - 10y - 19} \\ y(-2) &= 1 \end{align}$$ 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. $$ (-30xy - 16x - 10y - 13) dx + (-15x^2 - 10x - 10y - 19) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(-30xy - 16x - 10y - 13\right) = -30x - 10 = \frac{\partial}{\partial x}\left(-15x^2 - 10x - 10y - 19\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= -30xy - 16x - 10y - 13\\ \frac{\partial F}{\partial y} &= -15x^2 - 10x - 10y - 19 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (-30xy - 16x - 10y - 13)\,\partial x = -15x^2y - 8x^2 - 10xy - 13x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (-15x^2 - 10x - 10y - 19)\,\partial y = -15x^2y - 10xy - 5y^2 - 19y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = -5y^2 - 19y$ and $ \tilde{C}(x) = -8x^2 - 13x. $ So $$ F(x,y) = -15x^2y - 10xy - 8x^2 - 5y^2 - 13x - 19y. $$

  7. The solution is $F(x,y) = K.$ $$ -15x^2y - 10xy - 8x^2 - 5y^2 - 13x - 19y = K $$
Now we plug in the initial values $x = -2$ and $y = 1$ and solve for $K = -70$. So the solution to the initial value problem is $$ -15x^2y - 10xy - 8x^2 - 5y^2 - 13x - 19y = -70 $$ You may reload this page to generate additional examples.

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