Kansas State University

Mathematics Department

Textbook Contents

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{-3y^2 - y - 4x + 3}{6xy + x + 36y + 2} \\ y(2) &= 0 \end{align}$$ 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. $$ (3y^2 + y + 4x - 3) dx + (6xy + x + 36y + 2) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(3y^2 + y + 4x - 3\right) = 6y + 1 = \frac{\partial}{\partial x}\left(6xy + x + 36y + 2\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= 3y^2 + y + 4x - 3\\ \frac{\partial F}{\partial y} &= 6xy + x + 36y + 2 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (3y^2 + y + 4x - 3)\,\partial x = 3xy^2 + xy + 2x^2 - 3x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (6xy + x + 36y + 2)\,\partial y = 3xy^2 + xy + 18y^2 + 2y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = 18y^2 + 2y$ and $ \tilde{C}(x) = 2x^2 - 3x. $ So $$ F(x,y) = 3xy^2 + xy + 2x^2 + 18y^2 - 3x + 2y. $$

  7. The solution is $F(x,y) = K.$ $$ 3xy^2 + xy + 2x^2 + 18y^2 - 3x + 2y = K $$
Now we plug in the initial values $x = 2$ and $y = 0$ and solve for $K = 2$. So the solution to the initial value problem is $$ 3xy^2 + xy + 2x^2 + 18y^2 - 3x + 2y = 2 $$ You may reload this page to generate additional examples.

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