please remove my name
a good user was posting some good questions..........so here is one from my side.....very easy...[3][3]
let f be twice diff real valued function satisfying
f(x)+f"(x)=-x.g(x).f'(x) wheere g(x)≥0 for all real x.
Prove that mod(f(x)) is bounded
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9 Answers
consider
y = f(x)2+f'(x)2
now y≥0
also y' = 2f'(x)((f(x)+f"(x))=-x.g(x).f'(x)2 = -x X something non negative
observe y' changes sign only at x=0 and by direction of change of sign we can conclude y has maximum at x=0 only ( y = constant is also possible if gx is always 0 )
hence as y has maxima at 0 only and always greater than 0 its bounded
y bounded implys f(x)2 is bounded implying mod (fx) is also bounded
eureka u need to confirm ans abv
and it took more than 2 lines !!!!!!!!
i donno how guys like u and rohan call certain questions very easy [3]
well its just one liner becoz imp step is getting that function y. ..rest is mind game...u did almost 99% way i did....i just cut it short in the end.[3][3]
here is mine:
f(x)+f"(x)=-x.g(x).f'(x) ------------(1)
multiply both sides of eqn (1) by 2f'(x)
=> 2f(x).f'(x)+2f'(x).f"(x)=-2x.g(x).f'(x)2
=> d/dx (f(x)2+f'(x)2 )=-2x.g(x).f'(x)2
on observation, LHS>0 for x<0 and LHS<0 for x>0
=>function f(x)2+f'(x)2 increases to a maximum till x=0 and decreases after that.
=>f(x) and f'(x) are bounded.
hence proved.
This method was given in the book.....involves physics in calculus[3][3]
If we analyze the problem,we get to know that its physical significance is that f(x) is the amplitude of an oscillator with time dependent dampening x.g(x) .Since dampening is negative for x<o and posiitve for x>0
=>oscillator will gain energy before time t=0 and loses its remaining energy thereafter.
=>Bounded
dude that explanation given in book isnt so enlightening
how are we supposed to know dampening theorys and moreover dampening theory would ve been derived from the above explanations only !!!!
ya thats confusing explaiation....just posted to show how the authors show off their knowledge with useless things...