Kansas State University

Mathematics Department

Textbook Contents

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{-12y^2 - 4y + 2x - 6}{24xy + 4x + 28y + 2} \\ y(-2) &= 1 \end{align}$$ 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. $$ (12y^2 + 4y - 2x + 6) dx + (24xy + 4x + 28y + 2) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(12y^2 + 4y - 2x + 6\right) = 24y + 4 = \frac{\partial}{\partial x}\left(24xy + 4x + 28y + 2\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= 12y^2 + 4y - 2x + 6\\ \frac{\partial F}{\partial y} &= 24xy + 4x + 28y + 2 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (12y^2 + 4y - 2x + 6)\,\partial x = 12xy^2 + 4xy - x^2 + 6x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (24xy + 4x + 28y + 2)\,\partial y = 12xy^2 + 4xy + 14y^2 + 2y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = 14y^2 + 2y$ and $ \tilde{C}(x) = -x^2 + 6x. $ So $$ F(x,y) = 12xy^2 + 4xy - x^2 + 14y^2 + 6x + 2y. $$

  7. The solution is $F(x,y) = K.$ $$ 12xy^2 + 4xy - x^2 + 14y^2 + 6x + 2y = K $$
Now we plug in the initial values $x = -2$ and $y = 1$ and solve for $K = -32$. So the solution to the initial value problem is $$ 12xy^2 + 4xy - x^2 + 14y^2 + 6x + 2y = -32 $$ You may reload this page to generate additional examples.

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