The value of λ for which the integral equation 

\(y(x)=\lambda \displaystyle \int_0^1 x^2 e^{x+t} y(t) d t\)

has a non-zero solution, is

1
\(\frac{4}{1+e^2}\)
2
\(\frac{2}{1+e^2}\)
3
\(\frac{4}{e^2-1}\)
4
\(\frac{2}{e^2-1}\)

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