Answer
$X''(x) - \lambda xX(x) = 0$ and $T'(t) + \lambda tT(t) = 0$.
Work Step by Step
Given $tu_{xx} + xu_{t} = 0$.
Suppose, for the sake of argument, that $u(x,t) = X(x)T (t)$. Then
$t(X(x)T(t))_{xx} + x(x(X(x)T(t))_{t} = 0$,
$\frac{X''(x)}{xX(x)} = -\frac{T'(t)}{tT(t)} = \lambda$. So,
$X''(x) - \lambda x X(x) = 0$ and $T'(t) + tT(t) = 0$.
