Kansas State University

Mathematics Department

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{4y^2 - 4y - 8x - 3}{6y^2 - 8xy - 20y + 4x + 15} \\ y(-1) &= 1 \end{align}$$ 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. $$ (-4y^2 + 4y + 8x + 3) dx + (6y^2 - 8xy - 20y + 4x + 15) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(-4y^2 + 4y + 8x + 3\right) = -8y + 4 = \frac{\partial}{\partial x}\left(6y^2 - 8xy - 20y + 4x + 15\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= -4y^2 + 4y + 8x + 3\\ \frac{\partial F}{\partial y} &= 6y^2 - 8xy - 20y + 4x + 15 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (-4y^2 + 4y + 8x + 3)\,\partial x = -4xy^2 + 4xy + 4x^2 + 3x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (6y^2 - 8xy - 20y + 4x + 15)\,\partial y = 2y^3 - 4xy^2 - 10y^2 + 4xy + 15y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = 2y^3 - 10y^2 + 15y$ and $ \tilde{C}(x) = 4x^2 + 3x. $ So $$ F(x,y) = 2y^3 - 4xy^2 + 4xy + 4x^2 - 10y^2 + 3x + 15y. $$

  7. The solution is $F(x,y) = K.$ $$ 2y^3 - 4xy^2 + 4xy + 4x^2 - 10y^2 + 3x + 15y = K $$
Now we plug in the initial values $x = -1$ and $y = 1$ and solve for $K = 8$. So the solution to the initial value problem is $$ 2y^3 - 4xy^2 + 4xy + 4x^2 - 10y^2 + 3x + 15y = 8 $$ You may reload this page to generate additional examples.

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