i) x(0)=0, x(π/2)=0 
ii) x(0)=0, x(π/8)=0 
iii) x(0)=0, x(π/2)=1

in Calculus Answers by

Your answer

Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
To avoid this verification in future, please log in or register.

1 Answer

x"+16x=0 can be written x"=-16x which is classic simple harmonic motion, like a pendulum or oscillating spring. Its solution is a wave, a sine wave or cosine wave.

If x=Asin(nt+c)+Bcos(nt+c), x'=nAcos(nt+c)-nBsin(nt+c) and x"=-n^2Asin(nt+c)-n^2Bcos(nt+c)=-n^2x.

So if we put n^2x=16x, n=4 and x(t)=Asin(4t+c)+Bcos(4t+c).

(i) x(0)=0, x(π/2)=0: Asin(c)+Bcos(c)=0, tan(c)=-B/A;

(ii) x(0)=0, x(π/8)=0: Asin(c+π/2)+Bcos(c+π/2)=Acos(c)-Bsin(c)=0, tan(c)=A/B;

(iii) x(0)=0, x(π/2)=1: Asin(c)+Bcos(c)=1.

Another approach is to use the characteristic equation which gives us x=Ae^4it+Be^-4it where i=√-1.

e^iy=cos(y)+isin(y) and e^-iy=cos(y)-isin(y) and

e^iny=cos(ny)+isin(ny)=(cos(y)+isin(y))^n.

Therefore, x=A(cos(4t)+isin(4t))+B(cos(4t)-isin(4t)).

(i) x(0)=0, x(π/2)=0: 0=A+B, B=-A, x=2Aisin(4t)

(ii) x(0)=0, x(π/8)=0; 0=A+B, 0=iA-iB, A=B=0, x=0

(iii) x(0)=0, x(π/2)=1; 0=A+B, 1=A+B, which is inconsistent.

Should the question have x' instead of x in the second of the pair of boundary conditions?

by Top Rated User (642k points)

Related questions

1 answer
asked Sep 17, 2013 in Calculus Answers by mermaid_raid Level 1 User (120 points) | 168 views
0 answers
Welcome to MathHomeworkAnswers.org, where students, teachers and math enthusiasts can ask and answer any math question. Get help and answers to any math problem including algebra, trigonometry, geometry, calculus, trigonometry, fractions, solving expression, simplifying expressions and more. Get answers to math questions. Help is always 100% free!
82,892 questions
87,496 answers
1,965 comments
3,942 users