Question

If $f\left(x\right)=e^x+{\int }_0^1(e^x+t{e}^{-x})f\left(t\right)dt$, then prove that $f\left(x\right)=\frac{2(e-1)}{4e-2e^2}.e^x+\frac{e-1}{4-2e}.e^{-x}$.

Answer 1

We can write $f\left(x\right)=A{e}^x+B{e}^{-x}$, where
$A=1+{\int }_0^{1}f\left(t\right)dt$ and $B={\int }_0^{1}tf\left(t\right)dt$
$\therefore A=1+{\int }_0^1(A{e}^t+B{e}^{-t})dt$
$={(A{e}^t+B{e}^{-t})}_0^t$
$A=1+A(e^1-1)-B(e^{-1}-1)$
$⟹(2-e)A+(e^{-1}-1)B=1$ (i)
$B={\int }_0^{1}t(A{e}^t+B{e}^{-t})dt$
$=A{(t{e}^t-e^t)}_0^t+B{(-t{e}^{-t}-e^{-t})}_0^1$
$B=A+B(1-2e^{-1})$
$⟹A-2e^{-1}B=0$ (ii)
From equations (i) and (ii), we get
$A=\frac{2(e-1)}{4e-2e^2}$, $B=\frac{e-1}{4-2e}$
Hence, $f\left(x\right)=\frac{2(e-1)}{4e-2e^2}.e^x+\frac{e-1}{4-2e}.e^{-x}$