Kansas State University

Mathematics Department

Textbook Contents

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{8xy + 18x - 10y - 30}{-15y^2 - 4x^2 - 4y + 10x + 5} \\ y(-2) &= -1 \end{align}$$ 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. $$ (-8xy - 18x + 10y + 30) dx + (-15y^2 - 4x^2 - 4y + 10x + 5) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(-8xy - 18x + 10y + 30\right) = -8x + 10 = \frac{\partial}{\partial x}\left(-15y^2 - 4x^2 - 4y + 10x + 5\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= -8xy - 18x + 10y + 30\\ \frac{\partial F}{\partial y} &= -15y^2 - 4x^2 - 4y + 10x + 5 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (-8xy - 18x + 10y + 30)\,\partial x = -4x^2y - 9x^2 + 10xy + 30x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (-15y^2 - 4x^2 - 4y + 10x + 5)\,\partial y = -5y^3 - 4x^2y - 2y^2 + 10xy + 5y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = -5y^3 - 2y^2 + 5y$ and $ \tilde{C}(x) = -9x^2 + 30x. $ So $$ F(x,y) = -5y^3 - 4x^2y + 10xy - 9x^2 - 2y^2 + 30x + 5y. $$

  7. The solution is $F(x,y) = K.$ $$ -5y^3 - 4x^2y + 10xy - 9x^2 - 2y^2 + 30x + 5y = K $$
Now we plug in the initial values $x = -2$ and $y = -1$ and solve for $K = -62$. So the solution to the initial value problem is $$ -5y^3 - 4x^2y + 10xy - 9x^2 - 2y^2 + 30x + 5y = -62 $$ You may reload this page to generate additional examples.

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