Kansas State University

Mathematics Department

Textbook Contents

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Exact Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{-2y - 10x}{15y^2 + 8y + 2x + 6} \\ y(2) &= -1 \end{align}$$ 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. $$ (2y + 10x) dx + (15y^2 + 8y + 2x + 6) dy = 0 $$
  2. We test for exactness. $$\frac{\partial}{\partial y}\left(2y + 10x\right) = 2 = \frac{\partial}{\partial x}\left(15y^2 + 8y + 2x + 6\right) $$ so the equation is exact.

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

    Matching the expressions up, we find $C(y) = 5y^3 + 4y^2 + 6y$ and $ \tilde{C}(x) = 5x^2. $ So $$ F(x,y) = 5y^3 + 2xy + 5x^2 + 4y^2 + 6y. $$

  7. The solution is $F(x,y) = K.$ $$ 5y^3 + 2xy + 5x^2 + 4y^2 + 6y = K $$
Now we plug in the initial values $x = 2$ and $y = -1$ and solve for $K = 9$. So the solution to the initial value problem is $$ 5y^3 + 2xy + 5x^2 + 4y^2 + 6y = 9 $$ You may reload this page to generate additional examples.

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