Kansas State University

Mathematics Department

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Exact Equations

Additional Examples

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

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

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

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

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