Kansas State University

Mathematics Department

Textbook Contents

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{20y^2 + 5y + 8x - 15}{6y^2 - 40xy - 24y - 5x - 7} \\ y(2) &= 4 \end{align}$$ 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. $$ (-20y^2 - 5y - 8x + 15) dx + (6y^2 - 40xy - 24y - 5x - 7) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(-20y^2 - 5y - 8x + 15\right) = -40y - 5 = \frac{\partial}{\partial x}\left(6y^2 - 40xy - 24y - 5x - 7\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= -20y^2 - 5y - 8x + 15\\ \frac{\partial F}{\partial y} &= 6y^2 - 40xy - 24y - 5x - 7 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (-20y^2 - 5y - 8x + 15)\,\partial x = -20xy^2 - 5xy - 4x^2 + 15x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (6y^2 - 40xy - 24y - 5x - 7)\,\partial y = 2y^3 - 20xy^2 - 12y^2 - 5xy - 7y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = 2y^3 - 12y^2 - 7y$ and $ \tilde{C}(x) = -4x^2 + 15x. $ So $$ F(x,y) = 2y^3 - 20xy^2 - 5xy - 4x^2 - 12y^2 + 15x - 7y. $$

  7. The solution is $F(x,y) = K.$ $$ 2y^3 - 20xy^2 - 5xy - 4x^2 - 12y^2 + 15x - 7y = K $$
Now we plug in the initial values $x = 2$ and $y = 4$ and solve for $K = -758$. So the solution to the initial value problem is $$ 2y^3 - 20xy^2 - 5xy - 4x^2 - 12y^2 + 15x - 7y = -758 $$ You may reload this page to generate additional examples.

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