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Book
Advanced Engineering Mathematics
Author
Erwin Kreyszig
Edition
8th Edition
t = var('t') x = function('x',t) DE = diff(x, t, 2)*diff(x,t) -1 m, n = desolve(DE, [x,t]) #print m #print n #only taking m as one of the solutions, we will calculate the constants k1 and k2, using initial conditions. 'a' is the conditions where x=0 at t=0 a = m.subs(t=0) #print a b = diff(m, t).subs(t=0) #print b k1 = var('k1') b = function('b',k1) k1 = solve(b==2,k1) #print k1 a.subs(k1=2) k2 = var('k2') a = function('a',k2) k2 = solve(a==2,k2) print k2 x = m.subs(k1=2,k2=2,t=6) v = diff(m,t).subs(k1 =2, k2=2, t=6) print x print v #distance = 58/3m and velocity = 4m/s and the negative sign is just because of the direction. #k1 = k2 = 2 are the constants
Solution by:
- Zubin Mehta, Student(just graduated), IIT Bombay