= Solution
Set $T=t+a>0$. The <similarity solution> uses $y=(x+b)T^{-1/2}$, whose <partial derivatives> at fixed $x$ or $t$ are
$$
y_x=T^{-1/2},\qquad y_t=-\frac y{2T}.
$$
Apply the <multivariable chain rule> to $u=T^{-1/2}v(y)$:
$$
u_t=-\frac12T^{-3/2}v+T^{-1/2}v'\left(-\frac y{2T}\right)
=-\frac12T^{-3/2}(v+yv'),
$$
$$
u_x=T^{-1}v',\qquad u_{xx}=T^{-3/2}v''.
$$
Substituting in the <heat equation> and cancelling the positive common factor $T^{-3/2}$ gives
$$
\boxed{-\frac12(v+yv')=\lambda v''.}
$$
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