Kansas State University

Mathematics Department

Textbook Contents

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{-20y - 6}{9y^2 + 10y + 20x - 19} \\ y(-3) &= -5 \end{align}$$ 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. $$ (20y + 6) dx + (9y^2 + 10y + 20x - 19) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(20y + 6\right) = 20 = \frac{\partial}{\partial x}\left(9y^2 + 10y + 20x - 19\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= 20y + 6\\ \frac{\partial F}{\partial y} &= 9y^2 + 10y + 20x - 19 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (20y + 6)\,\partial x = 20xy + 6x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (9y^2 + 10y + 20x - 19)\,\partial y = 3y^3 + 5y^2 + 20xy - 19y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = 3y^3 + 5y^2 - 19y$ and $ \tilde{C}(x) = 6x. $ So $$ F(x,y) = 3y^3 + 20xy + 5y^2 + 6x - 19y. $$

  7. The solution is $F(x,y) = K.$ $$ 3y^3 + 20xy + 5y^2 + 6x - 19y = K $$
Now we plug in the initial values $x = -3$ and $y = -5$ and solve for $K = 127$. So the solution to the initial value problem is $$ 3y^3 + 20xy + 5y^2 + 6x - 19y = 127 $$ You may reload this page to generate additional examples.

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