Kansas State University

Mathematics Department

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{-4xy - 16x - 3y - 11}{2x^2 + 3x + 2y - 3} \\ y(1) &= -5 \end{align}$$ 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. $$ (4xy + 16x + 3y + 11) dx + (2x^2 + 3x + 2y - 3) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(4xy + 16x + 3y + 11\right) = 4x + 3 = \frac{\partial}{\partial x}\left(2x^2 + 3x + 2y - 3\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= 4xy + 16x + 3y + 11\\ \frac{\partial F}{\partial y} &= 2x^2 + 3x + 2y - 3 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (4xy + 16x + 3y + 11)\,\partial x = 2x^2y + 8x^2 + 3xy + 11x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (2x^2 + 3x + 2y - 3)\,\partial y = 2x^2y + 3xy + y^2 - 3y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = y^2 - 3y$ and $ \tilde{C}(x) = 8x^2 + 11x. $ So $$ F(x,y) = 2x^2y + 3xy + 8x^2 + y^2 + 11x - 3y. $$

  7. The solution is $F(x,y) = K.$ $$ 2x^2y + 3xy + 8x^2 + y^2 + 11x - 3y = K $$
Now we plug in the initial values $x = 1$ and $y = -5$ and solve for $K = 34$. So the solution to the initial value problem is $$ 2x^2y + 3xy + 8x^2 + y^2 + 11x - 3y = 34 $$ You may reload this page to generate additional examples.

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