Consider the 1-dimensional heat equation:
$$\left\{ \begin{align} & {{u}_{t}}\left( x,t \right)={{u}_{xx}}\left( x,t \right),\text{ }x\in R,\text{ }t>0 \\ & u\left( x,0 \right)={{e}^{a{{x}^{2}}}},\text{ }x\in R \\ \end{align} \right.$$
Find an explicit solution without integral signs.
I have tried separation of variables, Green's function, and Fourier transform but just couldn't resolve the integral because of the term ${{e}^{a{{x}^{2}}}}$. Please help.
Inspired by Heat equation with initial value answered by Mercy King.
To simplify, let $a=1$.
Setting $$\xi=\sqrt{\frac{1-4t}{4t}}\left(y-\frac{x}{1-4t}\right),$$ we have: \begin{eqnarray} \frac{(x-y)^2}{4t}-y^2&=&\frac{(1-4t)y^2-2xy+x^2}{4t}\\ &=&\frac{1-4t}{4t}\left[y^2-\frac{2x}{1-4t}y+\frac{x^2}{1-4t}\right]\\ &=&\frac{1-4t}{4t}\left[\left(y-\frac{x}{1-4t}\right)^2+\frac{x^2}{1-4t}-\frac{x^2}{(1-4t)^2}\right]\\ &=&\frac{1-4t}{4t}\left[\left(y-\frac{x}{1-4t}\right)^2-\frac{4x^2t}{(1-4t)^2}\right]\\ &=&\xi^2-\frac{x^2}{(1-4t)}. \end{eqnarray}
It follows that \begin{eqnarray} u(x,t)&=&\frac{1}{\sqrt{4\pi t}}\int_{-\infty}^\infty\exp\left(-\frac{(x-y)^2}{4t}+y^2\right)\, dy\\ &=&\frac{1}{\sqrt{4\pi t}}\sqrt{\frac{4t}{1-4t}}\exp[\frac{x^2}{(1-4t)}]\int_{-\infty}^\infty\exp(-\xi^2)\, d\xi\\ &=&\frac{1}{\sqrt{\pi(1-4t)}}\exp[\frac{x^2}{(1-4t)}]\int_{-\infty}^\infty\exp(-\xi^2)\, d\xi. \end{eqnarray} Using the fact that $$\int_{-\infty}^\infty\exp(-\xi^2)\, d\xi=\sqrt{\pi},$$ we get $$u(x,t)=\frac{1}{\sqrt{1-4t}}\exp\left(\frac{x^2}{(1-4t)}\right).$$