This preview shows pages 1–7. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Math 205, Spring 2011
B. Dodson
Text  GoodeAnnin, 3rd edt.
Oﬃce hours: M,F: 2:203:30, Tue: 11:3012:30
and by appt. email: bad0@lehigh.edu 2 1. Course Info
2. Homework 1:
Chapter 1, Sections 1, 2, 3, 4, 5, 6
Due Monday, Jan 24th Chapter 1: Intro, Slope Fields, verify solution
(Monday, 1/17)
1.4, 1.5: Separable DE, Population/logistic growth
(start 1.6; Wednesday, 1/19)
1.6 Linear Equations (Friday, 1/21 4 Problem 1:
√
Verify that the function y = c1 x is a
y
solution of y ′ =
2x
Solution:
Compute y ′ and check.
y′ = 1
1
−2
c1 ( 2 )x . √
c1 x
y
=
2x
2x
√
x
1
= c1 ( 2 ) √ 2
( x)
1
1
= c1 ( 2 ) √
x
= y ′. Problem 2:
Determine all values r so y = erx is a
solution to y ′′ − 4y ′ + 3y = 0. 6 Linear DE, Main Step: to solve linear DE
dy
+ p(x)y = q (x)
dx
multiply by the integral factor f = e p dx and use
d
(f y ) = f (y ′ + py )
dx
d
on the left to get
(f y ) = f q.
dx
Problem:
Solve dy
2x
+
y = 4x, −1 ≤ x ≤ 1.
2)
dx (1 − x Solution:
Find integral factor, inside integral ﬁrst:
2x
dx
2)
(1 − x = − ln(1 − x2 ) = ln (1 − x2 )−1 (simplify!). So e 2x
1−x2 )
( dx =e 2 −1 = 1−x Multiply by f =
1
1 − x2 ln ((1−x2 )−1 ) 1
=
= f.
1 − x2
1
and use Main Property:
2
1−x 2x
dy
+
y
2)
dx (1 − x = 4x
,
2
1−x 4x
.
2
1−x
y
= (−2 ln (1 − x2 )) + c,
Now integration gives
1 − x2
d
dx y
1 − x2 = so y = (1 − x2 ) − ln ((1 − x2 )2 ) + c . Notice that we can check this instance of the
Main Property directly (using the product rule). ...
View
Full
Document
This note was uploaded on 11/24/2011 for the course MATH 205 taught by Professor Zhang during the Spring '08 term at Lehigh University .
 Spring '08
 zhang
 Slope

Click to edit the document details