Kansas State University

Mathematics Department

Textbook Contents

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{2y^2 - 4y + 8}{-4xy + 4x - 28y + 25} \\ y(-3) &= -5 \end{align}$$ 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. $$ (-2y^2 + 4y - 8) dx + (-4xy + 4x - 28y + 25) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(-2y^2 + 4y - 8\right) = -4y + 4 = \frac{\partial}{\partial x}\left(-4xy + 4x - 28y + 25\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= -2y^2 + 4y - 8\\ \frac{\partial F}{\partial y} &= -4xy + 4x - 28y + 25 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (-2y^2 + 4y - 8)\,\partial x = -2xy^2 + 4xy - 8x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (-4xy + 4x - 28y + 25)\,\partial y = -2xy^2 + 4xy - 14y^2 + 25y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = -14y^2 + 25y$ and $ \tilde{C}(x) = -8x. $ So $$ F(x,y) = -2xy^2 + 4xy - 14y^2 - 8x + 25y. $$

  7. The solution is $F(x,y) = K.$ $$ -2xy^2 + 4xy - 14y^2 - 8x + 25y = K $$
Now we plug in the initial values $x = -3$ and $y = -5$ and solve for $K = -241$. So the solution to the initial value problem is $$ -2xy^2 + 4xy - 14y^2 - 8x + 25y = -241 $$ You may reload this page to generate additional examples.

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