Kansas State University

Mathematics Department

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Exact Equations

Additional Examples

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

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

    Matching the expressions up, we find $C(y) = 5y^3 + 9y^2 + 5y$ and $ \tilde{C}(x) = -2x^2 + 2x. $ So $$ F(x,y) = 5y^3 + 15xy^2 + 15xy - 2x^2 + 9y^2 + 2x + 5y. $$

  7. The solution is $F(x,y) = K.$ $$ 5y^3 + 15xy^2 + 15xy - 2x^2 + 9y^2 + 2x + 5y = K $$
Now we plug in the initial values $x = 0$ and $y = -1$ and solve for $K = -1$. So the solution to the initial value problem is $$ 5y^3 + 15xy^2 + 15xy - 2x^2 + 9y^2 + 2x + 5y = -1 $$ You may reload this page to generate additional examples.

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