Kansas State University

Mathematics Department

Textbook Contents

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{12xy + 20x + 3y + 3}{-6x^2 - 3x - 8y + 14} \\ y(0) &= 1 \end{align}$$ 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. $$ (-12xy - 20x - 3y - 3) dx + (-6x^2 - 3x - 8y + 14) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(-12xy - 20x - 3y - 3\right) = -12x - 3 = \frac{\partial}{\partial x}\left(-6x^2 - 3x - 8y + 14\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= -12xy - 20x - 3y - 3\\ \frac{\partial F}{\partial y} &= -6x^2 - 3x - 8y + 14 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (-12xy - 20x - 3y - 3)\,\partial x = -6x^2y - 10x^2 - 3xy - 3x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (-6x^2 - 3x - 8y + 14)\,\partial y = -6x^2y - 3xy - 4y^2 + 14y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = -4y^2 + 14y$ and $ \tilde{C}(x) = -10x^2 - 3x. $ So $$ F(x,y) = -6x^2y - 3xy - 10x^2 - 4y^2 - 3x + 14y. $$

  7. The solution is $F(x,y) = K.$ $$ -6x^2y - 3xy - 10x^2 - 4y^2 - 3x + 14y = K $$
Now we plug in the initial values $x = 0$ and $y = 1$ and solve for $K = 10$. So the solution to the initial value problem is $$ -6x^2y - 3xy - 10x^2 - 4y^2 - 3x + 14y = 10 $$ You may reload this page to generate additional examples.

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