Kansas State University

Mathematics Department

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{9y^2 - 12y - 10}{-15y^2 - 18xy + 16y + 12x - 14} \\ y(0) &= 1 \end{align}$$ 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. $$ (-9y^2 + 12y + 10) dx + (-15y^2 - 18xy + 16y + 12x - 14) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(-9y^2 + 12y + 10\right) = -18y + 12 = \frac{\partial}{\partial x}\left(-15y^2 - 18xy + 16y + 12x - 14\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= -9y^2 + 12y + 10\\ \frac{\partial F}{\partial y} &= -15y^2 - 18xy + 16y + 12x - 14 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (-9y^2 + 12y + 10)\,\partial x = -9xy^2 + 12xy + 10x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (-15y^2 - 18xy + 16y + 12x - 14)\,\partial y = -5y^3 - 9xy^2 + 8y^2 + 12xy - 14y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = -5y^3 + 8y^2 - 14y$ and $ \tilde{C}(x) = 10x. $ So $$ F(x,y) = -9xy^2 - 5y^3 + 12xy + 8y^2 + 10x - 14y. $$

  7. The solution is $F(x,y) = K.$ $$ -9xy^2 - 5y^3 + 12xy + 8y^2 + 10x - 14y = K $$
Now we plug in the initial values $x = 0$ and $y = 1$ and solve for $K = -11$. So the solution to the initial value problem is $$ -9xy^2 - 5y^3 + 12xy + 8y^2 + 10x - 14y = -11 $$ You may reload this page to generate additional examples.

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