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..

 

[CellCoree]

There are infinitely many real valued continuously differentiable functions $g$ on the real line that satisfy the given equation. One possible approach to finding these functions is to use the fundamental theorem of calculus to rewrite the integral on the right side of the equation as the difference between two antiderivatives of the integrand. This will result in a differential equation that can be solved to find the desired functions. Alternatively, one can also use a substitution such as $u=g(x)$ to transform the equation into a separable differential equation, which can then be solved using standard methods. Ultimately, the solution set will depend on the initial conditions and any additional constraints given in the problem.