The heat equation
$u_t=u_{xx}$
has the following symmetries (infinitesimal transformations):
$X_1 = \partial_x \quad X_2 = \partial_t \quad X_3 = u\partial_u \quad X_4 = x\partial_x +2t\partial_t$
$X_5 = 2t\partial_x-xu\partial_u \quad X_6 = tx\partial_x+t^2\partial_t-\frac{1}{4}(x^2+2t)u\partial_u$
I have some problems with operator $X_6$. We have 3 Lie's equations
1)$\frac{dt'}{da}=t'^2, \qquad t'(a=0)=t$
2)$\frac{dx'}{da}=x't', \qquad x'(a=0)=x$
3)$\frac{du'}{da}=-\frac{1}{4}(x'^2+2t')u', \quad u'(a=0)=u$
From the first we have $t'=\frac{t}{1-at}$, from the second $x'=\frac{x}{1-at}$, from the third $u'=u{e}^{-{\frac {axt}{-4\,a{t}^{2}+4\,t}}}\sqrt {-at+1}$.
And I'm having trouble with checking that
$u'_{t'} - u'_{x'x'}=0$
I assume that it is necessary to calculate
$\frac{\partial u'}{\partial t'} = \frac{\partial u'}{\partial t}\frac{\partial t}{\partial t'} + \frac{\partial u'}{\partial x}\frac{\partial x}{\partial t'}$
$\frac{\partial u'}{\partial x'} = \frac{\partial u'}{\partial x}\frac{\partial x}{\partial x'} + \frac{\partial u'}{\partial t}\frac{\partial t}{\partial x'}$
$\frac{\partial^2 u'}{\partial x'^2} = \frac{\partial \left(\frac{\partial u'}{\partial x'}\right)}{\partial x}\frac{\partial x}{\partial x'} + \frac{\partial \left(\frac{\partial u'}{\partial x'}\right)}{\partial t}\frac{\partial t}{\partial x'}$
but I don't get $u'_{t'} - u'_{x'x'}=0$
Thanks for your time.