Solution sets/ existence and uniqueness of solutions to $Ku-\lambda u=\int^1_0 \frac{x^2}{1+y^3}u(y)dy-\lambda u(x)=f(x)$ Given 
$$ Ku-\lambda u=\int^1_0 \frac{x^2}{1+y^3}u(y)dy-\lambda u(x)=f(x) $$
A) For what values of $\lambda$ does there exist a unique solution for all $f\in L^2(0,1)$?
B)  Find the solution set for all values of $\lambda$ and any $f\in L^2(0,1)$
My solution is below.
 A: Here are my results without the rigorous details.  Let me know if you come up with different results. Also, I'm nbt sure about the solution set for $\lambda=0$.
Consider $\lambda\neq 0$,
We note that $K$ is Hilbert-Schmidt so $-\lambda(I+(-\lambda^{-1}K))$ is Fredholm of index $0$.  Thus the existence of a solution is equivalent to its uniqueness.  
So we find $-\lambda(I+(-\lambda^{-1}K))u=0$ implies either $\lambda=\ln(2)/3$ with nullspace $\mathcal L (x^2 )$ or there exists a unique solution defined by 
$$ u(x)= \lambda^{-1}\left ((\ln(2)/(3\lambda)-1)^{-1}\left( \int^1_0 \frac{f(y)}{1+y^3}dy \right)-f(x) \right)$$
If $\lambda=\ln(2)/3$, then we find there exists a solution if $f(x)\perp (1+y^3)$, which leads to the solution set 
$$u(x)=3f(x)/\ln(2)+cx^2$$
for any $c$.  
Now consider $\lambda=0$,
$$ x^2\int^1_0 \frac{u(y)}{1+y^3}dy=f(x) $$
Here is my thought:  If $f(x)=cx^2$ for some $c$, then the solution set is $u(x)=c(1+x^3)+u_n$ where $u_n$ is any function such that $\int^1_0\frac{u_n(y)}{1+y^3}dy=0$.  
