# Download e-book for kindle: An Introduction to Ordinary Differential Equations (Dover by Earl A. Coddington

By Earl A. Coddington

ISBN-10: 0486659429

ISBN-13: 9780486659428

"Written in an admirably cleancut and cost-effective style." — *Mathematical Review*. an intensive, systematic first path in straightforward differential equations for undergraduates in arithmetic and technological know-how, requiring merely simple calculus for a history, and together with many routines designed to improve students' approach in fixing equations. With difficulties and solutions. Index.

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**Additional resources for An Introduction to Ordinary Differential Equations (Dover Books on Mathematics)**

**Example text**

3. 3. 4. 4 · · · (2n − 2) 1! 4 · · · (2n − 4) 2! 3 · · · (2n−1) n n(n−1) n−2 n(n−1)(n−2)(n−3) n−4 = x x + −· · · x − n! 4 = Pn (x). 12) we shall prove the following recurrence relation. 3 (Recurrence Relation). (n + 1)Pn+1 (x) = (2n + 1)xPn (x) − nPn−1 (x), n = 1, 2, · · · . 13) Proof. 12) with respect to t, we get (x − t)(1 − 2xt + t2 )−3/2 = ∞ nPn (x)tn−1 n=1 and hence (x − t)(1 − 2xt + t2 )−1/2 = (1 − 2xt + t2 ) ∞ nPn (x)tn−1 , n=1 which is the same as (x − t) ∞ n=0 Pn (x)tn = (1 − 2xt + t2 ) ∞ nPn (x)tn−1 .

N! 9) Proof. Let v = (x2 − 1)n , then (x2 − 1) dv = 2nxv. 10), (n + 1) times by Leibniz’s rule, we obtain (x2 −1) dn+2 v dn+1 v dn v dn+1 v dn v +2(n+1)x +n(n+1) = 2n x + (n+1) , dxn+2 dxn+1 dxn dxn+1 dxn 50 Lecture 7 which is the same as (1 − x2 ) dn+2 v dn+1 v dn v − 2x n+1 + n(n + 1) n = 0. 19) with a = n. Thus, it is necessary that z= dn v = cPn (x), dxn where c is a constant. Since Pn (1) = 1, we have c = dn v dxn n = k=0 = x=1 dn 2 (x − 1)n dxn = x=1 dn (x − 1)n (x + 1)n dxn n n! n! (x − 1)n−k (x + 1)k k (n − k)!

Once the particular form of the solution is known its construction is almost routine. 18). 10), where the functions p(x) and q(x) are analytic at x = x0 . 10) reduces to y ′′ + p(x) ′ q(x) y + 2 y = 0. 1) In comparison with at an ordinary point, the construction of a series solution at a singular point is diﬃcult. 3) exists in the interval J = (0, ∞). 2) can be represented by a power series with x = 0 as its point of expansion in any ∞ interval of the form (0, a), a > 0. 3) has this property. , y(x) = c1 x + c2 x2 is the general solution of x2 y ′′ − 2xy ′ + 2y = 0).

### An Introduction to Ordinary Differential Equations (Dover Books on Mathematics) by Earl A. Coddington

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