Kansas State University

Mathematics Department

Textbook Contents

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{20xy + 8x - 10y - 9}{-10x^2 + 10x + 4y + 27} \\ y(0) &= 3 \end{align}$$ 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. $$ (-20xy - 8x + 10y + 9) dx + (-10x^2 + 10x + 4y + 27) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(-20xy - 8x + 10y + 9\right) = -20x + 10 = \frac{\partial}{\partial x}\left(-10x^2 + 10x + 4y + 27\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= -20xy - 8x + 10y + 9\\ \frac{\partial F}{\partial y} &= -10x^2 + 10x + 4y + 27 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (-20xy - 8x + 10y + 9)\,\partial x = -10x^2y - 4x^2 + 10xy + 9x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (-10x^2 + 10x + 4y + 27)\,\partial y = -10x^2y + 10xy + 2y^2 + 27y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = 2y^2 + 27y$ and $ \tilde{C}(x) = -4x^2 + 9x. $ So $$ F(x,y) = -10x^2y + 10xy - 4x^2 + 2y^2 + 9x + 27y. $$

  7. The solution is $F(x,y) = K.$ $$ -10x^2y + 10xy - 4x^2 + 2y^2 + 9x + 27y = K $$
Now we plug in the initial values $x = 0$ and $y = 3$ and solve for $K = 99$. So the solution to the initial value problem is $$ -10x^2y + 10xy - 4x^2 + 2y^2 + 9x + 27y = 99 $$ You may reload this page to generate additional examples.

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