Solve integral equation $(f(x))^{2}=2 \int_{0}^{x} f(t) \sqrt{1-t^{2}} \mathrm{~d} t, \quad x \in[0,1]$
Solution 1:
At the stage where you have $2yy′=2y\sqrt{1−x^2},$ notice that $y=0$ is a solution. So the solutions are $f(x)=0$ and the solution you found for some fixed C. You should go and check which values of C makes the equation true. After doing the substitution into the original equation to find the allowed values of $C=y(0)=f(0),$ I found that $f(0)^2=\frac18.$ Hence, you actually have three solutions on $[0,1],$ but whoever assigned you the exercise may omit the trivial $f(x)=0.$