I have the integral equation $$u(x)=1+\lambda \int_0^1 K(x,t)u(t)dt $$ $x \in (0,1)$, $\lambda \in \mathbb{R} $ and $$K(x,t)=\begin{cases} x(t+1) & t \leq x \\ t(x+1) & x \leq t \end{cases} $$ Find the eigenvalues and eigenfunctions of the equation by first converting it into a boundary value problem.
I was able to convert the integral equation into an ODE, by splitting the integral from $0$ to $x$ and $x$ to $1$ i.e. $$u(x)=1+\lambda \int_0^x x(t+1)udt +\lambda \int_x^1 t(x+1)udt$$ and differentiating twice, which gave me the equation $$u''(x)=\lambda u(x)$$
I then attempted to find the boundary values, I have $$u(0)=1+\lambda \int_0^1 tu(t)dt $$ $$u(1)=1+\lambda \int_0^1 udt+\lambda \int_0^1 tudt$$ $$u'(0)=0$$ $$u'(1)=\lambda \int_0^1 udt$$ Those values are related by $$u(1)=u'(1)+u(0)$$
So the boundary values are $$u(1)=u'(1)+u(0) \ \ (1)$$ and $$u'(0)=0 \ \ (2) $$
Now to find the eigenvalues I solve $$u''=\lambda u$$ The solution is $$u(x)=c_1e^{\sqrt{\lambda }x}+c_2e^{-\sqrt{\lambda }x } \ \ (3)$$
I now use the boundary values to find $c_1,c_2$ so $$u'(0)=0=c_1-c_2$$ $$c_1=c_2$$ So $$u(0)=2c_1$$ $$u(1)=c_1(e^{\sqrt{\lambda } } +e^{-\sqrt{\lambda }} )$$ and $$u'(1)=c_1(\sqrt{\lambda } e^{\sqrt{\lambda }}-\sqrt{\lambda } e^{-\sqrt{\lambda }}) $$ By plugging those values in $(1)$ I get $$c_1(e^{\sqrt{\lambda }}+e^{-\sqrt{\lambda }}-e^{\sqrt{\lambda }} \sqrt{\lambda } + e^{-\sqrt{\lambda }} \sqrt{\lambda}-2)=0$$ $c_1$ can't be $0$ otherwise we would have the trivial solution so it must be true that $$e^{\sqrt{\lambda }}+e^{-\sqrt{\lambda }}-e^{\sqrt{\lambda }} \sqrt{\lambda } + e^{-\sqrt{\lambda }} \sqrt{\lambda}-2=0$$ Which is only true if $\lambda=0$ therefore $\lambda=0$ is the eigenvalue and if I plug $\lambda=0$ in the general solution of the ODE ($(3)$) I get $$u(t)=2c_1$$ so the eigenfunctions are constant.
Is my solution correct?