Kansas State University

Mathematics Department

Textbook Contents

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{-20xy + 4x - 5y - 1}{10x^2 + 5x + 19} \\ y(0) &= -4 \end{align}$$ 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. $$ (20xy - 4x + 5y + 1) dx + (10x^2 + 5x + 19) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(20xy - 4x + 5y + 1\right) = 20x + 5 = \frac{\partial}{\partial x}\left(10x^2 + 5x + 19\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= 20xy - 4x + 5y + 1\\ \frac{\partial F}{\partial y} &= 10x^2 + 5x + 19 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (20xy - 4x + 5y + 1)\,\partial x = 10x^2y - 2x^2 + 5xy + x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (10x^2 + 5x + 19)\,\partial y = 10x^2y + 5xy + 19y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = 19y$ and $ \tilde{C}(x) = -2x^2 + x. $ So $$ F(x,y) = 10x^2y - 2x^2 + 5xy + x + 19y. $$

  7. The solution is $F(x,y) = K.$ $$ 10x^2y - 2x^2 + 5xy + x + 19y = K $$
Now we plug in the initial values $x = 0$ and $y = -4$ and solve for $K = -76$. So the solution to the initial value problem is $$ 10x^2y - 2x^2 + 5xy + x + 19y = -76 $$ You may reload this page to generate additional examples.

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