Kansas State University

Mathematics Department

Textbook Contents

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{2y^2 + 3y + 4}{15y^2 - 4xy - 10y - 3x - 13} \\ y(-3) &= 0 \end{align}$$ 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. $$ (-2y^2 - 3y - 4) dx + (15y^2 - 4xy - 10y - 3x - 13) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(-2y^2 - 3y - 4\right) = -4y - 3 = \frac{\partial}{\partial x}\left(15y^2 - 4xy - 10y - 3x - 13\right) $$ so the equation is exact.

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

    Matching the expressions up, we find $C(y) = 5y^3 - 5y^2 - 13y$ and $ \tilde{C}(x) = -4x. $ So $$ F(x,y) = -2xy^2 + 5y^3 - 3xy - 5y^2 - 4x - 13y. $$

  7. The solution is $F(x,y) = K.$ $$ -2xy^2 + 5y^3 - 3xy - 5y^2 - 4x - 13y = K $$
Now we plug in the initial values $x = -3$ and $y = 0$ and solve for $K = 12$. So the solution to the initial value problem is $$ -2xy^2 + 5y^3 - 3xy - 5y^2 - 4x - 13y = 12 $$ You may reload this page to generate additional examples.

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