This preview shows page 1. Sign up to view the full content.
Unformatted text preview: 2 = − 1 ± √ 3 i. Hence the roots of the auxiliary equation are r 1 = r 2 = − 1 − √ 3 i and r 3 = r 4 = − 1+ √ 3 i . Thus two linearly independent solutions are e − t cos( √ 3 t ) and e − t sin( √ 3 t ), and we get two more linearly independent by multiplying them by t . This gives a general solution of the form y ( t ) = ( c 1 + c 2 t ) e − t cos( √ 3 t ) + ( c 3 + c 4 t ) e − t sin( √ 3 t ) . 39. (a) Comparing given equation with the CauchyEuler equation (21) in general form, we conclude that a = 3, b = 11, and c = − 3. Thus, the substitution x = e t leads to the equation (22) in Problem 38 with these values of parameters. That is, a d 2 y dt 2 + ( b − a ) dy dt + cy = 0 ⇒ 3 d 2 y dt 2 + 8 dy dt − 3 y = 0 . 187...
View
Full
Document
This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

Click to edit the document details