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Differential Equation - Find the Constant.

  1. Nov 3, 2008 #1
    Hey,

    1. The problem statement, all variables and given/known data.
    Given,
    [tex]
    {f(0)} = {0}
    [/tex]
    [tex]
    {{{f}^{\prime}}{(0)}} = {0}
    [/tex]

    Find the constant [tex]{C}[/tex] for the following and justify,
    [tex]
    {{\frac {1}{2}}{{\left({f(x)}\right)}^{2}} + {{\left({{{f}^{\prime}}{\left({x}\right)}}\right)}^{2}}} = {C}
    [/tex]

    2. Relevant equations.
    Calculus.

    3. The attempt at a solution.
    This problem is take from the proof of another problem and I follow what they're doing in that proof all except these last lines,
    [​IMG]
    I don't get exactly how from: [tex]{{f(0)} = {0}}[/tex] and [tex]{{f^{\prime}{(0)}} = {0}}[/tex]; they're able to determine that the constant is zero ([tex]{{C} = {0}}[/tex]).
    How do they determine that?

    Thanks,

    -PFStudent
     
  2. jcsd
  3. Nov 3, 2008 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    They put x=0 to determine the constant C, didn't they? But perhaps I don't get the problem, since you didn't really state where all of this stuff is coming from.
     
  4. Nov 4, 2008 #3
    Hey,
    This problem is part of another problem I was trying to prove, the proof for that problem is below,
    [​IMG]

    I follow what they're doing in the proof all except these last lines,
    [​IMG]

    I don't get exactly how from: [tex]{{f(0)} = {0}}[/tex] and [tex]{{f^{\prime}{(0)}} = {0}}[/tex]; they're able to determine that the constant is zero ([tex]{{C} = {0}}[/tex]).
    How do they determine that?

    Thanks,

    -PFStudent
     
  5. Nov 4, 2008 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    I still don't get what you don't get. If f(x)^2+f'(x)^2=C is a CONSTANT, then f(x)^2+f'(x)^2 has the same value C for ALL values of x. One value of x is 0. So C=f(0)^2+f'(0)^2=0+0. ?????
     
  6. Nov 4, 2008 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    if
    [tex]{{\frac {1}{2}}{{\left({f(x)}\right)}^{2}} + {{\left({{{f}^{\prime}}{\left({x}\right)}}\right)} ^{2}}} = {C}[/tex]
    for all x, then, in particular, it is true for x= 0 Since f(x)= 0 and f'(x)= 0,
    [tex]{{\frac {1}{2}}{{\left({0}\right)}^{2}} + {{\left({{0}}\right)} ^{2}}} = {C}[/tex]
    so C must be 0. It's that simple.
     
  7. Nov 5, 2008 #6
    Hey,
    Ahh, right...that is right. That is, given any function of [itex]{x}[/itex] equal to a constant, than for all values of [itex]{x}[/itex] the result is the same constant.
    In other words,
    [tex]
    {g(x)} = {K}
    {,}
    {\textcolor{white}{.}}
    {\textcolor{white}{.}}
    \mbox{for all}
    {\textcolor{white}{.}}
    {x}
    [/tex]
    It was the "[tex]{\mbox{for all}{\textcolor{white}{.}}{x}}[/tex]" part that I just now realized, :redface:.
    Right, and that value of [itex]{C}[/itex] is valid for all values of [itex]{x}[/itex], I just realized that, :redface:.

    Thanks so much for the help: Dick and HallsofIvy.

    Thanks,

    -PFStudent
     
    Last edited: Nov 6, 2008
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