Exact Equations

**ragnar** · January 10th 2011, 08:20 PM

So I'm reading through Ordinary Differential Equations by Tenenbaum and Pollard, page 76, Example 9.5, and I find it bizarre. It says that $\cos y dx - (x \sin y - y^{2})dy = 0$ is exact on $\mathbb{R} \times \mathbb{R}$ which I readily believe. It then says we can integrate each part from $(0,0)$ to $(x,y)$ , i.e. $\int_{0}^{x}\cos y dx + \int_{0}^{y}y^{2} dy$ . The second integrand baffles me. The book hints that, because we're taking the integral with respect to $y$ at the point $(x_{0}, y_{0}) = (0,0)$ with $x$ fixed, we can therefore eliminate the $x \sin y$ . But one: Why? Two: If that's so, can't we do the same with the other one and have it be $\int_{0}^{x}(\cos (0))dx = x$ ?

**ragnar** · January 10th 2011, 08:24 PM

Sorry, just saw that the proof I skipped over because it seemed to assume too much familiarity on my part, gives an explanation of this.

Thread: Exact Equations

LinkBack

Thread Tools

Display

Exact Equations

Similar Math Help Forum Discussions

Second Order ODE: Exact Equations

Proving implicit equations are exact

ODE exact equations

Are these differential equations exact?

Exact Equations