Kansas State University

Mathematics Department

Textbook Contents

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{4y^2 + 20y + 12}{-9y^2 - 8xy - 10y - 20x - 7} \\ y(-1) &= 0 \end{align}$$ 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. $$ (-4y^2 - 20y - 12) dx + (-9y^2 - 8xy - 10y - 20x - 7) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(-4y^2 - 20y - 12\right) = -8y - 20 = \frac{\partial}{\partial x}\left(-9y^2 - 8xy - 10y - 20x - 7\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= -4y^2 - 20y - 12\\ \frac{\partial F}{\partial y} &= -9y^2 - 8xy - 10y - 20x - 7 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (-4y^2 - 20y - 12)\,\partial x = -4xy^2 - 20xy - 12x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (-9y^2 - 8xy - 10y - 20x - 7)\,\partial y = -3y^3 - 4xy^2 - 5y^2 - 20xy - 7y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = -3y^3 - 5y^2 - 7y$ and $ \tilde{C}(x) = -12x. $ So $$ F(x,y) = -4xy^2 - 3y^3 - 20xy - 5y^2 - 12x - 7y. $$

  7. The solution is $F(x,y) = K.$ $$ -4xy^2 - 3y^3 - 20xy - 5y^2 - 12x - 7y = K $$
Now we plug in the initial values $x = -1$ and $y = 0$ and solve for $K = 12$. So the solution to the initial value problem is $$ -4xy^2 - 3y^3 - 20xy - 5y^2 - 12x - 7y = 12 $$ You may reload this page to generate additional examples.

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