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

[anemone]

Find all real valued continuously differentiable functions $g$ on the real line such that for all $x$,

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

 

[mathmari]

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, $$g(x)=\pm \sqrt{1990}e^x$$

 

[chisigma]

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

$\chi$ $\sigma$

 

[mathmari]

[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..