The Question I have is: $$u_t(x, t) − ku_{xx}(x, t) = 0$$ $$∀x ∈ \mathbb{R}, t > 0$$
subject to-
$u(x, 0) = x^2 − 3x − 1$ $∀x ∈ \mathbb{R}.$
I started off with
$$u(x,t)=\int_{\mathbb{R}}e^\frac{-(x-y)^2}{4kt}(y^2-3y-1)dy$$
and then I made $σ=y-x$ therefore I get
$$u(x,t)=\frac{1}{\sqrt{4 \pi kt}}\int_{\mathbb{R}}e^\frac{-σ^2}{4kt}((σ+x)^2-3(σ+x)-1)dσ$$
that simplifies to:
$$u(x,t)=\frac{1}{\sqrt{4 \pi kt}}\int_{\mathbb{R}}e^\frac{-σ^2}{4kt}(σ^2+2σx+x^2-3σ-3x-1)dσ$$
But from here I'm not sure how to integrate the answer and get any further. Any help would be greatly appreciated
Substitute $$u(x,t)=v(t)+x^2-3x-1$$ For $v(t)$ we get ODE problem $$v'(t)=2k,\quad v(0)=0.$$ $\Rightarrow$ $v(t)=2kt$
Solution of Initial Value Problem is $$u(x,t)=2kt+x^2-3x-1$$