Homogeneous Integral Equations

Question

In Arfken (3rd ed) ex. 16.5.1 he derives the integral equation for a one dimensional linear oscillator that includes the Green function (eq. 16.148). This equation is a homogeneous integral equation. I know how to solve analytically the differential equation and I know what the solution is [the eigenvalue is $n^2π^2$ while $y = \sin(nπx)$]. My question is that how to I solve analytically the integral equation:

$$ y(x) = λ\left[ \int_0^x G_2(x,t)y(t)dt + \int_x^1 G_1(x,t)y(t)dt\right]$$

where $G_2(x,t) = t(1-x)$ whereas $G_1(x,t) =x(1-t)$.

Answer 1

$y(x)=\lambda\left(\int_0^xt(1-x)y(t)~dt+\int_x^1x(1-t)y(t)~dt\right)$

$y(x)=\lambda\left((1-x)\int_0^xty(t)~dt-x\int_1^x(1-t)y(t)~dt\right)$

$y'(x)=\lambda\left((1-x)xy(x)-\int_0^xty(t)~dt-x(1-x)y(x)-\int_1^x(1-t)y(t)~dt\right)$

$y'(x)=-\lambda\left(\int_0^xty(t)~dt+\int_1^x(1-t)y(t)~dt\right)$

$y''(x)=-\lambda(xy(x)+(1-x)y(x))$

$y''(x)=-\lambda y(x)$

$y''(x)+\lambda y(x)=0$

$y(x)=C_1\sin\sqrt\lambda x+C_2\cos\sqrt\lambda x$

$y(0)=0$ :

$C_2=0$

$\therefore y(x)=C_1\sin\sqrt\lambda x$

$y(1)=0$ :

$C_1=0$

$\therefore y(x)=0$