Kansas State University

Mathematics Department

Textbook Contents

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Exact Equations

Additional Examples

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

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= -4xy + 2x - 6y - 1\\ \frac{\partial F}{\partial y} &= -15y^2 - 2x^2 + 4y - 6x \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (-4xy + 2x - 6y - 1)\,\partial x = -2x^2y + x^2 - 6xy - x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (-15y^2 - 2x^2 + 4y - 6x)\,\partial y = -5y^3 - 2x^2y + 2y^2 - 6xy + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = -5y^3 + 2y^2$ and $ \tilde{C}(x) = x^2 - x. $ So $$ F(x,y) = -2x^2y - 5y^3 - 6xy + x^2 + 2y^2 - x. $$

  7. The solution is $F(x,y) = K.$ $$ -2x^2y - 5y^3 - 6xy + x^2 + 2y^2 - x = K $$
Now we plug in the initial values $x = -2$ and $y = 0$ and solve for $K = 6$. So the solution to the initial value problem is $$ -2x^2y - 5y^3 - 6xy + x^2 + 2y^2 - x = 6 $$ You may reload this page to generate additional examples.

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