Question

Question No. 5
If any two solutions of a linear homogeneous differential equations on the interval [a, b] are linearly dependent on this interval, then their Wronskian is

Answer
A. O Identically zero
B. O never zero
C. O equal to 1
D. O equal to -1​

Answer 1

Answer:

If y1(x)y1(x) and y2(x)y2(x) are two solutions of equation y′′+P(x)y′+Q(x)y=0y″+P(x)y′+Q(x)y=0 on an interval [a,b][a,b] and have a common zero in this interval, show that one is a constant multiple of the 1other.

Suppose the initial values of the solutions y1y1 and y2y2 are defined as follows : y1(to)=0,y′1(to)=c1y1(to)=0,y1′(to)=c1 and

y2(to)=0,y′2(to)=c2,y2(to)=0,y2′(to)=c2, then,

y1−y2y1−y2 is also a solution to the given differential equation which satisfies :

(y1−y2)(to)=0,(y1 −y2) ′(to)=c1−c2(y1−y2)(to)=0,(y1 −y2) ′(to)=c1−c2.

Since, all of y1 ,y2, y1−y2y1 ,y2, y1−y2 are solutions to he given differential equation, as per the uniqueness theorem, there exist unique curves which satisfy the above initial conditions.

Hence, I do not completely understand why y1y1, y2y2 must be a constant multiple of the other . I found an answer to this problem here. However, I do not understand the answer quite much.

Could someone please give an explanation to my confusion above.