Kansas State University

Mathematics Department

Textbook Contents

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{-2y^2 + 10y + 8}{4xy - 2y - 10x} \\ y(-3) &= 1 \end{align}$$ 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. $$ (2y^2 - 10y - 8) dx + (4xy - 2y - 10x) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(2y^2 - 10y - 8\right) = 4y - 10 = \frac{\partial}{\partial x}\left(4xy - 2y - 10x\right) $$ so the equation is exact.

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

    Matching the expressions up, we find $C(y) = -y^2$ and $ \tilde{C}(x) = -8x. $ So $$ F(x,y) = 2xy^2 - 10xy - y^2 - 8x. $$

  7. The solution is $F(x,y) = K.$ $$ 2xy^2 - 10xy - y^2 - 8x = K $$
Now we plug in the initial values $x = -3$ and $y = 1$ and solve for $K = 47$. So the solution to the initial value problem is $$ 2xy^2 - 10xy - y^2 - 8x = 47 $$ You may reload this page to generate additional examples.

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