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

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## 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 (660k points)

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