Suppose $f(x)$ and $g(x)$ are two continuous functions defined for $0 \leq x \leq 1$. Given, $f(x)=\

Question

Suppose $f(x)$ and $g(x)$ are two continuous functions defined for $0 \leq x \leq 1$. Given, $f(x)=\int_0^1 e^{x+t} \cdot f(t) d t$ and $g(x)=\int_0^1 e^{x+t} \cdot g(t) d t+x$. Then:

Professor PC Thomas Classes · Answer

Here, $f(x)=e^x \underbrace{\int_0^1 e^t \cdot f(t) d t}_{A(x y)}$    ......(i)
$f(x)=A e^x \Rightarrow f(t)=A e^t $
Where, $A=\int_0^1 e^t \cdot f(t) d t \Rightarrow A=\int_0^1 e^t \cdot A e^t d t ; A=A \int_0^1 e^{2 t} d t 
eq 0$ Hence, $f(x)=0 \Rightarrow f(1)=0$ Again, $g(x)=e^x \int_0^1 e^t g(t) d t+x$ 
$\Rightarrow g(x)=B e^x+x$ ....(ii)
$\Rightarrow g(t)=B e^t+t$
Where, $B=\int_0^1 e^t g(t) d t \Rightarrow B=\int_0^t e^t\left(B e^t+tight) d t \Rightarrow B=B \int_0^1 e^{2 t} d t+\int_0^1 e^t t d t$ But $\quad \int_0^1 e^{2 t} d t=\frac{1}{2}\left(e^2-1ight)$ and $\int_0^1 t e^t d t=1$ 
$ 	herefore \quad B=\frac{B}{2}\left(e^2-1ight)+1 \Rightarrow 2 B=B\left(e^2-1ight)+2 \Rightarrow 3 B=B e^2+2 \Rightarrow B=\frac{2}{3-e^2} $
 From Equation (ii), $g(x)=\left(\frac{2}{3-e^2}ight) e^x+x \Rightarrow g(0)=\frac{2}{3-e^2}$ Also, $f(0)=0$ 
$ \begin{aligned} & 	herefore g(0)-f(0)=\frac{2}{3-e^2}-0=\frac{2}{3-e^2} ; g(2)=\frac{2 e^2}{3-e^2}+2=\frac{6}{3-e^2} \ & 	herefore \frac{g(0)}{g(2)}=\frac{2}{3-e^2} \cdot \frac{3-e^2}{6}=\frac{1}{3} \end{aligned} $