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4

Good luck ! Problem 1 Problem 2 Problem 3

0

In this Mathematica code, please enter the following instead: soln = NDSolve[{7000*q''[t] + 20000*q[t] - 800*f''[t]*f[t] == 0, 896*f''[t] - 480*h''[t] + 7848*f[t] - 800*q''[t]*f[t] == 0, 900*h''[t] - 480*f''[t] + 20000*h[t] == 0, q[0] == 0.1, h[0] == 0.1, f[0] == 0.}, {q[t], f[t], h[t]}, ...

1

You are rediscovering the decoding algorithms for Bose-Chaudhuri-Hocquenghem (BCH) codes. For double- and triple- error-correcting codes, people have worked simpler algorithms that do not require matrix inversion as the Peterson algorithm that has been pointed out to you. See, for example, Polkinghorn, "Decoding of double and triple error correcting ...

3

Let $u=t-x$ , Then $x=t-u$ $\dfrac{dx}{dt}=1-\dfrac{du}{dt}$ $\therefore1-\dfrac{du}{dt}=\dfrac{A(1-t+u)}{t-t^2}-\dfrac{B(t-u)-C(t-u)^2}{(t-t^2)u}$ $u-u\dfrac{du}{dt}=\dfrac{Au}{t}+\dfrac{Au^2}{t-t^2}+\dfrac{Cu^2+(B-2Ct)u+Ct^2-Bt}{t-t^2}$ $u-u\dfrac{du}{dt}=\dfrac{Au}{t}+\dfrac{(A+C)u^2+(B-2Ct)u+Ct^2-Bt}{t-t^2}$ ...

2

I tried to use Mathematica to solve your problem and it solved pretty fast. Here is what I've done.

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