Simple differential equation?

let $A=\displaystyle \int^{2}_{0} y.dx$

$\large{\frac{dy}{y+A}=dx}$

its solution is

$\large{ln(y+A)=x+c}$

given $y(0)=1$ so $c=ln(1+A)$

$y=(1+A)e^{x}-A$

$A=\displaystyle \int^{2}_{0} y.dx$

$\large {A=\displaystyle \int^{2}_{0} ((1+A)e^{x}-A).dx}$

$\large{A=\dfrac{e^{2}-1}{4-e^2}}$

Hence solution is

$\huge{y(x)=\dfrac{3e^x-e^2+1}{4-e^2}}$