Kansas State University

Mathematics Department

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Exact Equations

Additional Examples

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

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

    Matching the expressions up, we find $C(y) = y^3 + 6y^2 + 3y$ and $ \tilde{C}(x) = -2x^2. $ So $$ F(x,y) = -10xy^2 + y^3 - 20xy - 2x^2 + 6y^2 + 3y. $$

  7. The solution is $F(x,y) = K.$ $$ -10xy^2 + y^3 - 20xy - 2x^2 + 6y^2 + 3y = K $$
Now we plug in the initial values $x = -2$ and $y = -5$ and solve for $K = 302$. So the solution to the initial value problem is $$ -10xy^2 + y^3 - 20xy - 2x^2 + 6y^2 + 3y = 302 $$ You may reload this page to generate additional examples.

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