Kansas State University

Mathematics Department

Textbook Contents

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{-6x - 6y + 3}{3y^2 - 4y + 6x + 1} \\ y(2) &= 3 \end{align}$$ 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. $$ (6x + 6y - 3) dx + (3y^2 - 4y + 6x + 1) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(6x + 6y - 3\right) = 6 = \frac{\partial}{\partial x}\left(3y^2 - 4y + 6x + 1\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= 6x + 6y - 3\\ \frac{\partial F}{\partial y} &= 3y^2 - 4y + 6x + 1 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (6x + 6y - 3)\,\partial x = 3x^2 + 6xy - 3x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (3y^2 - 4y + 6x + 1)\,\partial y = y^3 - 2y^2 + 6xy + y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

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

  7. The solution is $F(x,y) = K.$ $$ y^3 + 6xy + 3x^2 - 2y^2 - 3x + y = K $$
Now we plug in the initial values $x = 2$ and $y = 3$ and solve for $K = 54$. So the solution to the initial value problem is $$ y^3 + 6xy + 3x^2 - 2y^2 - 3x + y = 54 $$ You may reload this page to generate additional examples.

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