Kansas State University

Mathematics Department

Textbook Contents

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{5y^2 + 15y + 26}{-10xy - 15x - 8} \\ y(-2) &= -5 \end{align}$$ 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. $$ (-5y^2 - 15y - 26) dx + (-10xy - 15x - 8) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(-5y^2 - 15y - 26\right) = -10y - 15 = \frac{\partial}{\partial x}\left(-10xy - 15x - 8\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= -5y^2 - 15y - 26\\ \frac{\partial F}{\partial y} &= -10xy - 15x - 8 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (-5y^2 - 15y - 26)\,\partial x = -5xy^2 - 15xy - 26x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (-10xy - 15x - 8)\,\partial y = -5xy^2 - 15xy - 8y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = -8y$ and $ \tilde{C}(x) = -26x. $ So $$ F(x,y) = -5xy^2 - 15xy - 26x - 8y. $$

  7. The solution is $F(x,y) = K.$ $$ -5xy^2 - 15xy - 26x - 8y = K $$
Now we plug in the initial values $x = -2$ and $y = -5$ and solve for $K = 192$. So the solution to the initial value problem is $$ -5xy^2 - 15xy - 26x - 8y = 192 $$ You may reload this page to generate additional examples.

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