Kansas State University

Mathematics Department

Textbook Contents

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{20xy + 5y + 12x}{-3y^2 - 10x^2 + 6y - 5x - 2} \\ y(1) &= -4 \end{align}$$ 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. $$ (-20xy - 5y - 12x) dx + (-3y^2 - 10x^2 + 6y - 5x - 2) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(-20xy - 5y - 12x\right) = -20x - 5 = \frac{\partial}{\partial x}\left(-3y^2 - 10x^2 + 6y - 5x - 2\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= -20xy - 5y - 12x\\ \frac{\partial F}{\partial y} &= -3y^2 - 10x^2 + 6y - 5x - 2 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (-20xy - 5y - 12x)\,\partial x = -10x^2y - 5xy - 6x^2 + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (-3y^2 - 10x^2 + 6y - 5x - 2)\,\partial y = -y^3 - 10x^2y + 3y^2 - 5xy - 2y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = -y^3 + 3y^2 - 2y$ and $ \tilde{C}(x) = -6x^2. $ So $$ F(x,y) = -10x^2y - y^3 - 5xy - 6x^2 + 3y^2 - 2y. $$

  7. The solution is $F(x,y) = K.$ $$ -10x^2y - y^3 - 5xy - 6x^2 + 3y^2 - 2y = K $$
Now we plug in the initial values $x = 1$ and $y = -4$ and solve for $K = 174$. So the solution to the initial value problem is $$ -10x^2y - y^3 - 5xy - 6x^2 + 3y^2 - 2y = 174 $$ You may reload this page to generate additional examples.

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