By James C. Robinson

ISBN-10: 0511165986

ISBN-13: 9780511165986

ISBN-10: 0521533910

ISBN-13: 9780521533911

ISBN-10: 0521826500

ISBN-13: 9780521826501

This advent to dull differential and distinction equations is desirable not just for mathematicians yet for scientists and engineers in addition. certain ideas equipment and qualitative methods are lined, and plenty of illustrative examples are integrated. Matlab is used to generate graphical representations of strategies. a variety of routines are featured and proved strategies can be found for academics.

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**Additional resources for An Introduction to Ordinary Differential Equations **

**Sample text**

3 Find the function g(x) deﬁned for x > −1 that has slope ln(1 + x) and passes through the origin. 2)). It is not possible to ﬁnd explicit solutions of these equations in general. However, in certain cases the equations reduce to something much simpler. Suppose that a ﬂuid is ﬂowing down a pipe that has a circular cross-section of radius a. Assuming that the velocity V of the ﬂuid depends only on its distance from the centre of the pipe, the equation satisﬁed by V is 1 d r dr where P is a positive constant.

5. The stability or instability of a stationary point x ∗ can be determined from the value of f (x ∗ ) provided that f (x ∗ ) = 0. If f (x ∗ ) < 0 then the stationary point is stable, and if f (x ∗ ) > 0 the stationary point is unstable. 3 Analytic conditions for stability and instability There are very simple conditions on the derivative of f which will let us know whether a stationary point x ∗ is stable or unstable without having to sketch the graph of f . e. e. if f (x ∗ ) > 0, then the point will be unstable.

Y at some particular x value, x0 . 8). 8) that satisﬁes y(x0 ) = y0 . You should make sure that you understand what follows, since we will use similar reasoning very often throughout the rest of the book. For the ﬁrst method we do a little more than we have to: we ﬁnd the general solution, and then solve a very simple algebraic equation to ﬁnd the correct constant. 8) is y(x) = F(x) + c. The particular solution that we want has y(x0 ) = y0 , and so we need y0 = y(x0 ) = F(x0 ) + c ⇒ c = y0 − F(x0 ).

### An Introduction to Ordinary Differential Equations by James C. Robinson

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