Kansas State University

Mathematics Department

Textbook Contents

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{10xy + 4x - 20y - 7}{-5x^2 + 6y + 20x} \\ y(1) &= 3 \end{align}$$ 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. $$ (-10xy - 4x + 20y + 7) dx + (-5x^2 + 6y + 20x) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(-10xy - 4x + 20y + 7\right) = -10x + 20 = \frac{\partial}{\partial x}\left(-5x^2 + 6y + 20x\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= -10xy - 4x + 20y + 7\\ \frac{\partial F}{\partial y} &= -5x^2 + 6y + 20x \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (-10xy - 4x + 20y + 7)\,\partial x = -5x^2y - 2x^2 + 20xy + 7x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (-5x^2 + 6y + 20x)\,\partial y = -5x^2y + 3y^2 + 20xy + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

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

  7. The solution is $F(x,y) = K.$ $$ -5x^2y - 2x^2 + 20xy + 3y^2 + 7x = K $$
Now we plug in the initial values $x = 1$ and $y = 3$ and solve for $K = 77$. So the solution to the initial value problem is $$ -5x^2y - 2x^2 + 20xy + 3y^2 + 7x = 77 $$ You may reload this page to generate additional examples.

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