Kansas State University

Mathematics Department

Textbook Contents

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{4y^2 + 16y + 4x - 15}{-3y^2 - 8xy - 8y - 16x - 9} \\ y(-2) &= 0 \end{align}$$ 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. $$ (-4y^2 - 16y - 4x + 15) dx + (-3y^2 - 8xy - 8y - 16x - 9) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(-4y^2 - 16y - 4x + 15\right) = -8y - 16 = \frac{\partial}{\partial x}\left(-3y^2 - 8xy - 8y - 16x - 9\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= -4y^2 - 16y - 4x + 15\\ \frac{\partial F}{\partial y} &= -3y^2 - 8xy - 8y - 16x - 9 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (-4y^2 - 16y - 4x + 15)\,\partial x = -4xy^2 - 16xy - 2x^2 + 15x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (-3y^2 - 8xy - 8y - 16x - 9)\,\partial y = -y^3 - 4xy^2 - 4y^2 - 16xy - 9y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = -y^3 - 4y^2 - 9y$ and $ \tilde{C}(x) = -2x^2 + 15x. $ So $$ F(x,y) = -y^3 - 4xy^2 - 16xy - 2x^2 - 4y^2 + 15x - 9y. $$

  7. The solution is $F(x,y) = K.$ $$ -y^3 - 4xy^2 - 16xy - 2x^2 - 4y^2 + 15x - 9y = K $$
Now we plug in the initial values $x = -2$ and $y = 0$ and solve for $K = -38$. So the solution to the initial value problem is $$ -y^3 - 4xy^2 - 16xy - 2x^2 - 4y^2 + 15x - 9y = -38 $$ You may reload this page to generate additional examples.

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