Kansas State University

Mathematics Department

Textbook Contents

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{-5y^2 + 15y - 10x + 21}{10xy - 15x - 8} \\ y(-2) &= -4 \end{align}$$ 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. $$ (5y^2 - 15y + 10x - 21) dx + (10xy - 15x - 8) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(5y^2 - 15y + 10x - 21\right) = 10y - 15 = \frac{\partial}{\partial x}\left(10xy - 15x - 8\right) $$ so the equation is exact.

  3. Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= 5y^2 - 15y + 10x - 21\\ \frac{\partial F}{\partial y} &= 10xy - 15x - 8 \end{align}$$
  4. Integrate the first partial differential equation. $$ F(x,y) = \int (5y^2 - 15y + 10x - 21)\,\partial x = 5xy^2 - 15xy + 5x^2 - 21x + C(y) $$
  5. Integrate the second partial differential equation. $$ F(x,y) = \int (10xy - 15x - 8)\,\partial y = 5xy^2 - 15xy - 8y + \tilde{C}(x) $$
  6. Equate the expressions for F(x,y).

    Matching the expressions up, we find $C(y) = -8y$ and $ \tilde{C}(x) = 5x^2 - 21x. $ So $$ F(x,y) = 5xy^2 - 15xy + 5x^2 - 21x - 8y. $$

  7. The solution is $F(x,y) = K.$ $$ 5xy^2 - 15xy + 5x^2 - 21x - 8y = K $$
Now we plug in the initial values $x = -2$ and $y = -4$ and solve for $K = -186$. So the solution to the initial value problem is $$ 5xy^2 - 15xy + 5x^2 - 21x - 8y = -186 $$ You may reload this page to generate additional examples.

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