Kansas State University

Mathematics Department

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{-2y^2 + 10y - 6x - 4}{4xy - 10x + 4y - 5} \\ y(-3) &= -4 \end{align}$$ 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. $$ (2y^2 - 10y + 6x + 4) dx + (4xy - 10x + 4y - 5) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(2y^2 - 10y + 6x + 4\right) = 4y - 10 = \frac{\partial}{\partial x}\left(4xy - 10x + 4y - 5\right) $$ so the equation is exact.

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

    Matching the expressions up, we find $C(y) = 2y^2 - 5y$ and $ \tilde{C}(x) = 3x^2 + 4x. $ So $$ F(x,y) = 2xy^2 - 10xy + 3x^2 + 2y^2 + 4x - 5y. $$

  7. The solution is $F(x,y) = K.$ $$ 2xy^2 - 10xy + 3x^2 + 2y^2 + 4x - 5y = K $$
Now we plug in the initial values $x = -3$ and $y = -4$ and solve for $K = -149$. So the solution to the initial value problem is $$ 2xy^2 - 10xy + 3x^2 + 2y^2 + 4x - 5y = -149 $$ You may reload this page to generate additional examples.

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