Differential Equation - Find the Constant.

[PFStudent]

Hey,

1. Homework Statement .
Given,
$${f(0)} = {0}$$
$${{{f}^{\prime}}{(0)}} = {0}$$

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

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,
$${g(x)} = {K} {,} {\textcolor{white}{.}} {\textcolor{white}{.}} \mbox{for all} {\textcolor{white}{.}} {x}$$
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