Kansas State University

Mathematics Department

Textbook Contents

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{-40xy - 42x - 16y - 14}{20x^2 + 16x + 8y - 10} \\ y(0) &= -3 \end{align}$$ 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. $$ (40xy + 42x + 16y + 14) dx + (20x^2 + 16x + 8y - 10) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(40xy + 42x + 16y + 14\right) = 40x + 16 = \frac{\partial}{\partial x}\left(20x^2 + 16x + 8y - 10\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= 40xy + 42x + 16y + 14\\ \frac{\partial F}{\partial y} &= 20x^2 + 16x + 8y - 10 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (40xy + 42x + 16y + 14)\,\partial x = 20x^2y + 21x^2 + 16xy + 14x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (20x^2 + 16x + 8y - 10)\,\partial y = 20x^2y + 16xy + 4y^2 - 10y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = 4y^2 - 10y$ and $ \tilde{C}(x) = 21x^2 + 14x. $ So $$ F(x,y) = 20x^2y + 16xy + 21x^2 + 4y^2 + 14x - 10y. $$

  7. The solution is $F(x,y) = K.$ $$ 20x^2y + 16xy + 21x^2 + 4y^2 + 14x - 10y = K $$
Now we plug in the initial values $x = 0$ and $y = -3$ and solve for $K = 66$. So the solution to the initial value problem is $$ 20x^2y + 16xy + 21x^2 + 4y^2 + 14x - 10y = 66 $$ You may reload this page to generate additional examples.

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