Kansas State University

Mathematics Department

Textbook Contents

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{10y^2 + 25y - 9}{-3y^2 - 20xy + 12y - 25x + 5} \\ y(2) &= 3 \end{align}$$ 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. $$ (-10y^2 - 25y + 9) dx + (-3y^2 - 20xy + 12y - 25x + 5) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(-10y^2 - 25y + 9\right) = -20y - 25 = \frac{\partial}{\partial x}\left(-3y^2 - 20xy + 12y - 25x + 5\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= -10y^2 - 25y + 9\\ \frac{\partial F}{\partial y} &= -3y^2 - 20xy + 12y - 25x + 5 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (-10y^2 - 25y + 9)\,\partial x = -10xy^2 - 25xy + 9x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (-3y^2 - 20xy + 12y - 25x + 5)\,\partial y = -y^3 - 10xy^2 + 6y^2 - 25xy + 5y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = -y^3 + 6y^2 + 5y$ and $ \tilde{C}(x) = 9x. $ So $$ F(x,y) = -10xy^2 - y^3 - 25xy + 6y^2 + 9x + 5y. $$

  7. The solution is $F(x,y) = K.$ $$ -10xy^2 - y^3 - 25xy + 6y^2 + 9x + 5y = K $$
Now we plug in the initial values $x = 2$ and $y = 3$ and solve for $K = -270$. So the solution to the initial value problem is $$ -10xy^2 - y^3 - 25xy + 6y^2 + 9x + 5y = -270 $$ You may reload this page to generate additional examples.

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