Differential Equation - Find the Constant.

[PFStudent]

Hey,

1. Homework Statement .
Given,
<br /> {f(0)} = {0}<br />
<br /> {{{f}^{\prime}}{(0)}} = {0}<br />

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

2. Homework 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,
http://d.imagehost.org/0790/line.jpg
I don't get exactly how from: {{f(0)} = {0}} and {{f^{\prime}{(0)}} = {0}}; they're able to determine that the constant is zero ({{C} = {0}}).
How do they determine that?

Thanks,

-PFStudent

 

[Dick]

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.

 

[PFStudent]

Hey,

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.

This problem is part of another problem I was trying to prove, the proof for that problem is below,
http://d.imagehost.org/0948/Lemma_2_4.jpg

I follow what they're doing in the proof all except these last lines,
http://d.imagehost.org/0790/line.jpg

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

Thanks,

-PFStudent

 

[Dick]

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. ?

 

[HallsofIvy]

if
{{\frac {1}{2}}{{\left({f(x)}\right)}^{2}} + {{\left({{{f}^{\prime}}{\left({x}\right)}}\right)} ^{2}}} = {C}
for all x, then, in particular, it is true for x= 0 Since f(x)= 0 and f'(x)= 0,
{{\frac {1}{2}}{{\left({0}\right)}^{2}} + {{\left({{0}}\right)} ^{2}}} = {C}
so C must be 0. It's that simple.

 

[PFStudent]

Hey,

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. ?

Ahh, right...that is right. That is, given any function of {x} equal to a constant, than for all values of {x} the result is the same constant.
In other words,
<br /> {g(x)} = {K}<br /> {,}<br /> {\textcolor{white}{.}}<br /> {\textcolor{white}{.}}<br /> \mbox{for all}<br /> {\textcolor{white}{.}}<br /> {x}<br />
It was the "{\mbox{for all}{\textcolor{white}{.}}{x}}" part that I just now realized, .

if
{{\frac {1}{2}}{{\left({f(x)}\right)}^{2}} + {{\left({{{f}^{\prime}}{\left({x}\right)}}\right)}^{2}}} = {C}
for all x, then, in particular, it is true for x= 0 Since f(x)= 0 and f'(x)= 0,
{{\frac{1}{2}}{{\left({0}\right)}^{2}} + {{\left({{0}}\right)}^{2}}} = {C}
so C must be 0. It's that simple.

Right, and that value of {C} is valid for all values of {x}, I just realized that, .

Thanks so much for the help: Dick and HallsofIvy.

Thanks,

-PFStudent