Kansas State University

Mathematics Department

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{30xy + 2x + 10y - 7}{3y^2 - 15x^2 - 4y - 10x - 13} \\ y(-1) &= -1 \end{align}$$ 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. $$ (-30xy - 2x - 10y + 7) dx + (3y^2 - 15x^2 - 4y - 10x - 13) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(-30xy - 2x - 10y + 7\right) = -30x - 10 = \frac{\partial}{\partial x}\left(3y^2 - 15x^2 - 4y - 10x - 13\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= -30xy - 2x - 10y + 7\\ \frac{\partial F}{\partial y} &= 3y^2 - 15x^2 - 4y - 10x - 13 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (-30xy - 2x - 10y + 7)\,\partial x = -15x^2y - x^2 - 10xy + 7x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (3y^2 - 15x^2 - 4y - 10x - 13)\,\partial y = y^3 - 15x^2y - 2y^2 - 10xy - 13y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = y^3 - 2y^2 - 13y$ and $ \tilde{C}(x) = -x^2 + 7x. $ So $$ F(x,y) = y^3 - 15x^2y - 10xy - x^2 - 2y^2 + 7x - 13y. $$

  7. The solution is $F(x,y) = K.$ $$ y^3 - 15x^2y - 10xy - x^2 - 2y^2 + 7x - 13y = K $$
Now we plug in the initial values $x = -1$ and $y = -1$ and solve for $K = 7$. So the solution to the initial value problem is $$ y^3 - 15x^2y - 10xy - x^2 - 2y^2 + 7x - 13y = 7 $$ You may reload this page to generate additional examples.

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