Suppose a continuous function $f:\left(0,\infty \right)\to R$ satisfies $f\left(x\right)=2{\int }_0^{x}tf\left(t\right)dt+1,\forall x\ge 0$. Then, $f\left(1\right)$ equals

It is given that a continuous function 

$f:[0,\infty )⟶R$ satisfies

$f\left(x\right)=2{\int }_0^{x}tf\left(t\right)dt+1,\forall x\ge 0$

$\therefore f^{\text{'}}\left(x\right)=2xf\left(x\right)$

$\Rightarrow \frac{f^{\text{'}}\left(x\right)}{f\left(x\right)}=2x$

On integrating both sides, we get 

In $\left|f\left(x\right)\right|=x^2+c$

$\Rightarrow f\left(x\right)=K{e}^{x^2}$, where $K=e^c$

$\because f\left(0\right)=2{\int }_0^{0}tf\left(t\right)dt+1=1$

$\therefore K=1$

$\therefore f\left(1\right)=e$