# 3.25 - Math 334 Lecture#11 3.2 Linear Second Order ODEs...

This preview shows pages 1–2. Sign up to view the full content.

Math 334 Lecture #11 § 3.2: Linear Second Order ODE’s, Part II Examples. (a) Recall that the functions y 1 = e - t and y 2 = e - 3 t for t R are solutions of y 00 + 4 y 0 + 3 y = 0. The Wronskian of these two solutions is W ( y 1 , y 2 )( t ) = ± ± ± ± e - t e - 3 t - e - t - 3 e - 3 t ± ± ± ± = - 3 e - 4 t + e - 4 t = - 2 e - 4 t 6 = 0 , so that y = c 1 e - t + c 2 e - 3 t is the general solution. (b) The function y 1 = x , y 2 = xe x are solutions of x 2 y 00 - x ( x + 2) y 0 + ( x + 2) y = 0 on the open interval x > 0, for which W ( y 1 , y 2 )( x ) = ± ± ± ± x xe x 1 e x + xe x ± ± ± ± = xe x + x 2 e x - xe x = x 2 e x 6 = 0 for x > 0 , so that y = c 1 x + c 2 xe x is the general solution. (c) The functions y 1 ( t ) = e 3 t and y 2 ( t ) = e 3( t - 2) are solutions of y 00 - 6 y 0 + 9 y = 0. Their Wronskian is W ( y 1 , y 2 )( t ) = ± ± ± ± e 3 t e 3( t - 2) 3 e 3 t 3 e 3( t - 2) ± ± ± ± = 3 e 6 t - 6 - 3 e 6 t - 6 = 0 for all t I. So y 1 and y 2 do not form a fundamental set of solutions. Could it be that the Wronskian of two solutions of a second order linear homogeneous

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 2

3.25 - Math 334 Lecture#11 3.2 Linear Second Order ODEs...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online