Kansas State University

Mathematics Department

Textbook Contents

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{8y^2 + 10y + 15}{-16xy - 10x - 38y - 25} \\ y(-1) &= 0 \end{align}$$ 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. $$ (-8y^2 - 10y - 15) dx + (-16xy - 10x - 38y - 25) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(-8y^2 - 10y - 15\right) = -16y - 10 = \frac{\partial}{\partial x}\left(-16xy - 10x - 38y - 25\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= -8y^2 - 10y - 15\\ \frac{\partial F}{\partial y} &= -16xy - 10x - 38y - 25 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (-8y^2 - 10y - 15)\,\partial x = -8xy^2 - 10xy - 15x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (-16xy - 10x - 38y - 25)\,\partial y = -8xy^2 - 10xy - 19y^2 - 25y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = -19y^2 - 25y$ and $ \tilde{C}(x) = -15x. $ So $$ F(x,y) = -8xy^2 - 10xy - 19y^2 - 15x - 25y. $$

  7. The solution is $F(x,y) = K.$ $$ -8xy^2 - 10xy - 19y^2 - 15x - 25y = K $$
Now we plug in the initial values $x = -1$ and $y = 0$ and solve for $K = 15$. So the solution to the initial value problem is $$ -8xy^2 - 10xy - 19y^2 - 15x - 25y = 15 $$ You may reload this page to generate additional examples.

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