Kansas State University

Mathematics Department

Textbook Contents

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{-8y^2 + 6y - 2x + 4}{16xy - 6x - 40y + 14} \\ y(0) &= -5 \end{align}$$ 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. $$ (8y^2 - 6y + 2x - 4) dx + (16xy - 6x - 40y + 14) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(8y^2 - 6y + 2x - 4\right) = 16y - 6 = \frac{\partial}{\partial x}\left(16xy - 6x - 40y + 14\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= 8y^2 - 6y + 2x - 4\\ \frac{\partial F}{\partial y} &= 16xy - 6x - 40y + 14 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (8y^2 - 6y + 2x - 4)\,\partial x = 8xy^2 - 6xy + x^2 - 4x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (16xy - 6x - 40y + 14)\,\partial y = 8xy^2 - 6xy - 20y^2 + 14y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = -20y^2 + 14y$ and $ \tilde{C}(x) = x^2 - 4x. $ So $$ F(x,y) = 8xy^2 - 6xy + x^2 - 20y^2 - 4x + 14y. $$

  7. The solution is $F(x,y) = K.$ $$ 8xy^2 - 6xy + x^2 - 20y^2 - 4x + 14y = K $$
Now we plug in the initial values $x = 0$ and $y = -5$ and solve for $K = -570$. So the solution to the initial value problem is $$ 8xy^2 - 6xy + x^2 - 20y^2 - 4x + 14y = -570 $$ You may reload this page to generate additional examples.

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