Kansas State University

Mathematics Department

Textbook Contents

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{15y^2 - 3y + 16}{-30xy + 3x + 32y - 9} \\ y(-1) &= -4 \end{align}$$ 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. $$ (-15y^2 + 3y - 16) dx + (-30xy + 3x + 32y - 9) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(-15y^2 + 3y - 16\right) = -30y + 3 = \frac{\partial}{\partial x}\left(-30xy + 3x + 32y - 9\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= -15y^2 + 3y - 16\\ \frac{\partial F}{\partial y} &= -30xy + 3x + 32y - 9 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (-15y^2 + 3y - 16)\,\partial x = -15xy^2 + 3xy - 16x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (-30xy + 3x + 32y - 9)\,\partial y = -15xy^2 + 3xy + 16y^2 - 9y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = 16y^2 - 9y$ and $ \tilde{C}(x) = -16x. $ So $$ F(x,y) = -15xy^2 + 3xy + 16y^2 - 16x - 9y. $$

  7. The solution is $F(x,y) = K.$ $$ -15xy^2 + 3xy + 16y^2 - 16x - 9y = K $$
Now we plug in the initial values $x = -1$ and $y = -4$ and solve for $K = 560$. So the solution to the initial value problem is $$ -15xy^2 + 3xy + 16y^2 - 16x - 9y = 560 $$ You may reload this page to generate additional examples.

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