Kansas State University

Mathematics Department

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{-4y^2 - 10y + 6x + 1}{3y^2 + 8xy + 18y + 10x + 14} \\ y(-2) &= -1 \end{align}$$ 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. $$ (4y^2 + 10y - 6x - 1) dx + (3y^2 + 8xy + 18y + 10x + 14) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(4y^2 + 10y - 6x - 1\right) = 8y + 10 = \frac{\partial}{\partial x}\left(3y^2 + 8xy + 18y + 10x + 14\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= 4y^2 + 10y - 6x - 1\\ \frac{\partial F}{\partial y} &= 3y^2 + 8xy + 18y + 10x + 14 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (4y^2 + 10y - 6x - 1)\,\partial x = 4xy^2 + 10xy - 3x^2 - x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (3y^2 + 8xy + 18y + 10x + 14)\,\partial y = y^3 + 4xy^2 + 9y^2 + 10xy + 14y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = y^3 + 9y^2 + 14y$ and $ \tilde{C}(x) = -3x^2 - x. $ So $$ F(x,y) = y^3 + 4xy^2 + 10xy - 3x^2 + 9y^2 - x + 14y. $$

  7. The solution is $F(x,y) = K.$ $$ y^3 + 4xy^2 + 10xy - 3x^2 + 9y^2 - x + 14y = K $$
Now we plug in the initial values $x = -2$ and $y = -1$ and solve for $K = -4$. So the solution to the initial value problem is $$ y^3 + 4xy^2 + 10xy - 3x^2 + 9y^2 - x + 14y = -4 $$ You may reload this page to generate additional examples.

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