Kansas State University

Mathematics Department

Textbook Contents

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{4xy + 20x - 5y - 15}{-2x^2 + 5x + 2y + 1} \\ y(0) &= -5 \end{align}$$ 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. $$ (-4xy - 20x + 5y + 15) dx + (-2x^2 + 5x + 2y + 1) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(-4xy - 20x + 5y + 15\right) = -4x + 5 = \frac{\partial}{\partial x}\left(-2x^2 + 5x + 2y + 1\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= -4xy - 20x + 5y + 15\\ \frac{\partial F}{\partial y} &= -2x^2 + 5x + 2y + 1 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (-4xy - 20x + 5y + 15)\,\partial x = -2x^2y - 10x^2 + 5xy + 15x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (-2x^2 + 5x + 2y + 1)\,\partial y = -2x^2y + 5xy + y^2 + y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = y^2 + y$ and $ \tilde{C}(x) = -10x^2 + 15x. $ So $$ F(x,y) = -2x^2y + 5xy - 10x^2 + y^2 + 15x + y. $$

  7. The solution is $F(x,y) = K.$ $$ -2x^2y + 5xy - 10x^2 + y^2 + 15x + y = K $$
Now we plug in the initial values $x = 0$ and $y = -5$ and solve for $K = 20$. So the solution to the initial value problem is $$ -2x^2y + 5xy - 10x^2 + y^2 + 15x + y = 20 $$ You may reload this page to generate additional examples.

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