MHB Find Real Valued Functions: g(x)^2=Int+(1990)

  • Thread starter Thread starter anemone
  • Start date Start date
  • Tags Tags
    Functions
AI Thread Summary
The discussion revolves around finding real-valued continuously differentiable functions \( g \) that satisfy the equation \( (g(x))^2 = \int_{0}^{x} [(g(t))^2 + (g'(t))^2]\,dt + 1990 \). Through differentiation, it is established that \( g'(x) = g(x) \), leading to the solution \( g(x) = c e^x \). Substituting this back into the original equation reveals that \( c^2 = 1990 \), resulting in \( c = \pm \sqrt{1990} \). Thus, the functions satisfying the equation are \( g(x) = \pm \sqrt{1990} e^x \). The conclusion confirms the existence of such functions.
anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
Find all real valued continuously differentiable functions $g$ on the real line such that for all $x$,

$\displaystyle (g(x))^2=\int_{0}^{x} [(g(t))^2+(g'(t))^2]\,dt+1990$
 
Mathematics news on Phys.org
anemone said:
Find all real valued continuously differentiable functions $g$ on the real line such that for all $x$,

$\displaystyle (g(x))^2=\int_{0}^{x} [(g(t))^2+(g'(t))^2]\,dt+1990$

$$\displaystyle (g(x))^2=\int_{0}^{x} [(g(t))^2+(g'(t))^2]\,dt+1990 \Rightarrow \left ((g(x))^2 \right )'=\left ( \int_{0}^{x} [(g(t))^2+(g'(t))^2]\,dt+1990 \right )' \Rightarrow \\ 2g(x)g'(x)=(g(x))^2+(g'(x))^2 \Rightarrow (g(x))^2-2g(x)g'(x)+(g'(x))^2=0 \Rightarrow \\ (g(x)-g'(x))^2=0 \Rightarrow g'(x)=g(x) \Rightarrow g(x)=ce^x$$

$\displaystyle (g(x))^2=\int_{0}^{x} [(g(t))^2+(g'(t))^2]\,dt+1990$

for $x=0$ we get:

$\displaystyle (g(0))^2=\int_{0}^{0} [(g(t))^2+(g'(t))^2]\,dt+1990=1990 \Rightarrow g(0)= \pm \sqrt{1990}$$\displaystyle{g(0)=ce^0 \Rightarrow c=\pm \sqrt{1990}}$

Therefore, $$g(x)=\pm \sqrt{1990}e^x$$
 
Last edited by a moderator:
anemone said:
Find all real valued continuously differentiable functions $g$ on the real line such that for all $x$,

$\displaystyle (g(x))^2=\int_{0}^{x} [(g(t))^2+(g'(t))^2]\,dt+1990$

[sp]Deriving both terms You obtain...

$\displaystyle 2\ g\ g^{\ '} = g^{2} + g^{\ '\ 2} \implies (g - g^{\ '})^{2} = 0 \implies g^{\ '} = g\ (1)$

... and (1) is an ODE the solution of which is $\displaystyle g(x)= c\ e^{x}$. But inserting that is the original equation You obtain the false identity $\displaystyle c^{2}\ e^{2 x} = c^{2}\ e^{2 x} + 1990$, so that it exists no g(x) satisfying the original equation...[/sp]

Kind regards

$\chi$ $\sigma$
 
chisigma said:
[sp]Deriving both terms You obtain...

$\displaystyle 2\ g\ g^{\ '} = g^{2} + g^{\ '\ 2} \implies (g - g^{\ '})^{2} = 0 \implies g^{\ '} = g\ (1)$

... and (1) is an ODE the solution of which is $\displaystyle g(x)= c\ e^{x}$. But inserting that is the original equation You obtain the false identity $\displaystyle c^{2}\ e^{2 x} = c^{2}\ e^{2 x} + 1990$, so that it exists no g(x) satisfying the original equation...[/sp]

Kind regards

$\chi$ $\sigma$

Inserting $g(x)= c\ e^{x}$ in the original equation we get:

$$\displaystyle (g(x))^2=\int_{0}^{x} [(g(t))^2+(g'(t))^2]\,dt+1990 \Rightarrow c^2 e^{2x}=\int_{0}^{x} [c^2e^{2t}+c^2e^{2t}]\,dt+1990 \\ \Rightarrow c^2 e^{2x}=\int_{0}^{x} [2c^2e^{2t}]\,dt+1990 \Rightarrow c^2 e^{2x}=c^2e^{2x}-c^2+1990 \Rightarrow c^2=1990 \Rightarrow c=\pm \sqrt{1990}$$

Therefore, it exists a $g(x)$ satisfying the original equaltion..
 
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...

Similar threads

Back
Top