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Give an interval over which the set of two functions ๐’‡๐Ÿ(๐’™)=๐’™๐Ÿ and ๐’‡๐Ÿ(๐’™)=๐’™|๐’™| is linearly independent. Then give an interval over which the set consisting of ๐’‡๐Ÿ and ๐’‡๐Ÿ is linearly dependent.

Short Answer

Expert verified

The interval over which the set consisting of F1 and F2 is:

Linearly independent on the interval (โˆ’โˆž,โˆž).

Linearly dependent on the interval (โˆ’โˆž,0) or (0,โˆž).

Step by step solution

01

Define

According to Theorem 4.1.3: Criterion for linearly independent Solutions, the set of two solutions ๐‘ฆ1 and ๐‘ฆ2 is linearly independent on an interval ๐ผ if and only if ๐‘Š(๐‘ฆ1,๐‘ฆ2) โ‰  0 for every ๐‘ฅ in the interval.

02

Solve the differential equations.

The function, ๐‘“2(๐‘ฅ)=๐‘ฅ|๐‘ฅ| can be broken down into two parts.

f2(x)=-x2for(-โˆž,0)x2for(0,โˆž)

Since ๐‘“2(๐‘ฅ) changes its sign at ๐‘ฅ=0 in the interval (โˆ’โˆž,โˆž), neither of functions ๐‘“1(๐‘ฅ)=๐‘ฅ2and ๐‘“2(๐‘ฅ)=๐‘ฅ|๐‘ฅ| can be written as a linear combination of the other, and therefore, we conclude that the set of functions is linearly independent on the interval (โˆ’โˆž,โˆž).

For the interval (โˆ’โˆž,0), the Wronskian is

๐‘Š(๐‘“1(๐‘ฅ),๐‘“2(๐‘ฅ))=๐‘Š(๐‘ฅ2,โˆ’๐‘ฅ2)=x2-x22x-2x=x2โ‹…(โˆ’2๐‘ฅ)โˆ’2๐‘ฅโ‹…(โˆ’๐‘ฅ2)=โˆ’2x3+2x3=0

Similarly, for the interval (0,โˆž), the Wronskian

๐‘Š(๐‘“1(๐‘ฅ), ๐‘“2(๐‘ฅ)) = ๐‘Š(๐‘ฅ2, ๐‘ฅ2) = 0

For both the intervals, since ๐‘Š(๐‘“1(๐‘ฅ), ๐‘“2(๐‘ฅ))=0 for all ๐‘ฅ, we conclude that the set of functions is linearly dependent on the interval (โˆ’โˆž,0) or (0,โˆž).

An interval over which the set consisting of ๐‘“1 and ๐‘“2 is linearly dependent is

Linearly independent on the interval (โˆ’โˆž,โˆž).

Linearly dependent on the interval (โˆ’โˆž,0) or (0,โˆž).

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