Kansas State University

Mathematics Department

Textbook Contents

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{-3y^2 - 5y - 10x - 6}{6xy + 5x - 26y - 15} \\ y(2) &= -3 \end{align}$$ 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. $$ (3y^2 + 5y + 10x + 6) dx + (6xy + 5x - 26y - 15) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(3y^2 + 5y + 10x + 6\right) = 6y + 5 = \frac{\partial}{\partial x}\left(6xy + 5x - 26y - 15\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= 3y^2 + 5y + 10x + 6\\ \frac{\partial F}{\partial y} &= 6xy + 5x - 26y - 15 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (3y^2 + 5y + 10x + 6)\,\partial x = 3xy^2 + 5xy + 5x^2 + 6x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (6xy + 5x - 26y - 15)\,\partial y = 3xy^2 + 5xy - 13y^2 - 15y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = -13y^2 - 15y$ and $ \tilde{C}(x) = 5x^2 + 6x. $ So $$ F(x,y) = 3xy^2 + 5xy + 5x^2 - 13y^2 + 6x - 15y. $$

  7. The solution is $F(x,y) = K.$ $$ 3xy^2 + 5xy + 5x^2 - 13y^2 + 6x - 15y = K $$
Now we plug in the initial values $x = 2$ and $y = -3$ and solve for $K = -16$. So the solution to the initial value problem is $$ 3xy^2 + 5xy + 5x^2 - 13y^2 + 6x - 15y = -16 $$ You may reload this page to generate additional examples.

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