Kansas State University

Mathematics Department

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{-10xy + 25y - 11}{-12y^2 + 5x^2 + 10y - 25x + 7} \\ y(-3) &= 2 \end{align}$$ 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. $$ (10xy - 25y + 11) dx + (-12y^2 + 5x^2 + 10y - 25x + 7) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(10xy - 25y + 11\right) = 10x - 25 = \frac{\partial}{\partial x}\left(-12y^2 + 5x^2 + 10y - 25x + 7\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= 10xy - 25y + 11\\ \frac{\partial F}{\partial y} &= -12y^2 + 5x^2 + 10y - 25x + 7 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (10xy - 25y + 11)\,\partial x = 5x^2y - 25xy + 11x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (-12y^2 + 5x^2 + 10y - 25x + 7)\,\partial y = -4y^3 + 5x^2y + 5y^2 - 25xy + 7y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = -4y^3 + 5y^2 + 7y$ and $ \tilde{C}(x) = 11x. $ So $$ F(x,y) = 5x^2y - 4y^3 - 25xy + 5y^2 + 11x + 7y. $$

  7. The solution is $F(x,y) = K.$ $$ 5x^2y - 4y^3 - 25xy + 5y^2 + 11x + 7y = K $$
Now we plug in the initial values $x = -3$ and $y = 2$ and solve for $K = 209$. So the solution to the initial value problem is $$ 5x^2y - 4y^3 - 25xy + 5y^2 + 11x + 7y = 209 $$ You may reload this page to generate additional examples.

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