Trouble with needless of absolute value

  • Thread starter mech-eng
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  • #1
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mary.png
Hi, while solving differential equations problems we have to sometimes use absolute value while
taking an integral of which 1 over a function. But in some problems I understand that we do not
have to use absolute value sign for example if the function related with atom number which can
never be a negative number we do not have to use absolute value. But in the picture I added here there is no sign about physical meaning of the problem so I think we should use absolute value and so there should have to be two different answers. The second picture of handwritting is
my other solution. The last one is a topic related with the problems. Have a nice day.
 

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  • #2
HallsofIvy
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Equation 2.8 certainly should be [itex]ln|y+ 1|= ln|x|+ cnst= ln|x|+ ln|a|= ln|ax|[/itex]

However, once we have gotten rid of the logarithms on both sides, we have y+ 1= ax, whether y, x, and a are positive or negative, and no longer need the absolute value signs.
 
  • #3
760
11
Equation 2.8 certainly should be [itex]ln|y+ 1|= ln|x|+ cnst= ln|x|+ ln|a|= ln|ax|[/itex]

However, once we have gotten rid of the logarithms on both sides, we have y+ 1= ax, whether y, x, and a are positive or negative, and no longer need the absolute value signs.
But in this situation y +1 might be positive whereas ax might negative so y + 1 might equal to -ax which leads to two different equalities, isn't it?
 
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  • #4
HallsofIvy
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No, that can't happen. "a" is a constant determined by the initial value. If we are given an initial value [itex](x_0, y_0)[/itex] such that [itex]y_0+ 1[/itex] and [itex]x_0[/itex] have the same sign, then a will be positive and y+ 1 and x will have the same sign for all x. If we are given an initial value [itex](x_0, y_0)[/itex] such that [itex]y_0+ 1[/itex] and [itex]x_0[/itex] have opposite sign, then a is negative and y+1 and x will have opposite sign for all x. In either case, y+ 1 and ax will have the same sign
 

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