Answered: A. {p(t) | √p(t)dt = 0} B. {p(t) | p′…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Related questions

Question

Let P2 denote the vector space of polynomials of degree up to 2. Which of the following subsets of P2 are subspaces of P2?

A. {p(t) | S₁¹ p(t)dt = 0}
B. {p(t) | p' (0) = p(7)}
C. {p(t) | p' (t) is constant}
D. {p(t) | p(4) = 3}
E. {p(t) | p' (t) + 4p(t) + 2 = 0}
F. {p(t) | p(8) = 0}

Expert Solution

Step 1: Problem (A)

Theorem: The necessary and sufficient condition for a sub set W of a vector soace V(F) is $a. alpha space plus space b. beta space element of space W$

for every $alpha comma space beta space element of space W space a n d space a comma space b space element of space F$

(A) Given sub set of $P subscript 2$ is $W space equals space open curly brackets p open parentheses t close parentheses space divided by space integral subscript 0 superscript 1 p open parentheses t close parentheses d t space equals space 0 close curly brackets$

Let $p open parentheses t close parentheses comma space q open parentheses t close parentheses space element of space W space rightwards double arrow space$ $integral subscript 0 superscript 1 p open parentheses t close parentheses d t space equals space 0 space space space a n d space integral subscript 0 superscript 1 q open parentheses t close parentheses d t space equals space 0$

Let $a comma space b space element of space R$ ( scalars in field )

consider $integral subscript 0 superscript 1 open square brackets a. p open parentheses t close parentheses plus b. q open parentheses t close parentheses close square brackets d t space equals space a integral subscript 0 superscript 1 p open parentheses t close parentheses d t space plus b integral subscript 0 superscript 1 q open parentheses t close parentheses d t space$

$equals space a space open parentheses 0 close parentheses space plus space b open parentheses 0 close parentheses$

$equals space 0$

From the definition of given set W, $a. p open parentheses t close parentheses plus b. q open parentheses t close parentheses space element of space W$ $for all space p open parentheses t close parentheses comma space q open parentheses t close parentheses space element of space W space a n d space a comma space b space element of space R$

therefore W is a subspace of $P subscript 2$