For what value(/s) of $\lambda$ , solution of the following Integral Equation does not exist ?$$y(x)=1+\lambda\int_0^1(1-3xt)y(t)\,dt$$
Let , $$y(x)=1+\lambda C_1-3\lambda xC_2$$where , $$C_1=\int_0^1(1+\lambda C_1-3\lambda tC_2)\,dt=1+\lambda C_1-\frac{3}{2}\lambda C_2$$
$$\implies(1-\lambda )C_1+\frac{3}{2}\lambda C_2=1$$
and $$C_2=\int_0^1t(1+\lambda C_1-3\lambda tC_2)\,dt=\frac{1}{2}+\frac{\lambda C_1}{2}-\lambda C_2$$
$$\implies \frac{\lambda C_1}{2}-(1+\lambda)C_2=-\frac{1}{2}$$
Now , $$D(\lambda)=\left|\begin{matrix}1-\lambda & \frac{3\lambda}{2}\\\frac{\lambda}{2} & -(1+\lambda)\end{matrix}\right|$$We know if $D(\lambda)\not=0$ then the I.E. has unique solution and if $D(\lambda)=0$ then the I.E. has either no solution or infinitely many solutions.
Now $D(\lambda)=0$ gives $\lambda=\pm 2$.For , $\lambda=2$ we can not eliminate $C_1 ,C_2$. So in this case the I.E. has infinitely many solutions.
When $\lambda=-2$ then we can find $C_1$ and $C_2$..
But I can not find the values of $\lambda$ for which solution does not exist
When $\lambda =2$, we find that the system of equations
$$-C_1+3C_2=1$$
and
$$C_1-3C_2=-\frac12$$
has no solution. Thus, the integral equation has no solution for $\lambda =2$.
When $\lambda =2$, we find that the system of equations
$$3C_1-3C_2=1$$
and
$$-C_1+C_2=-\frac12$$
has no solution. Thus, the integral equation has no solution for $\lambda =-2$.