Kansas State University

Mathematics Department

Textbook Contents

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{-20xy + 6x - 10y + 7}{10x^2 + 10x - 7} \\ y(0) &= 0 \end{align}$$ 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. $$ (20xy - 6x + 10y - 7) dx + (10x^2 + 10x - 7) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(20xy - 6x + 10y - 7\right) = 20x + 10 = \frac{\partial}{\partial x}\left(10x^2 + 10x - 7\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= 20xy - 6x + 10y - 7\\ \frac{\partial F}{\partial y} &= 10x^2 + 10x - 7 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (20xy - 6x + 10y - 7)\,\partial x = 10x^2y - 3x^2 + 10xy - 7x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (10x^2 + 10x - 7)\,\partial y = 10x^2y + 10xy - 7y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = -7y$ and $ \tilde{C}(x) = -3x^2 - 7x. $ So $$ F(x,y) = 10x^2y - 3x^2 + 10xy - 7x - 7y. $$

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

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