I have the following equation: $u_t - u_{xx} = 0$ with initial data $u(x, 0) = e^{kx}$ for some constant $k$ and $x \in \mathbb{R}, t > 0$ I'm looking for the general solution $u(x, t)$.
So far, I have set up the solution as follows:
$$u(x, t) = \frac{1}{2\sqrt{\pi t}} \int_{-\infty}^{\infty} e^{\frac{-(x - y)^2}{4t}} e^{ky} dy$$
But I ran in to trouble evaluating it. Any help would be much appreciated.
On
You can directly solve the integral by completing the square: $$ -(y-x)^2 + Ky = -\big(y-x-\frac K2\big)^2+Kx+\frac{K^2}4.$$ E.g. use $a^2 - b^2 = (a-b)(a+b)$ with $a-b=\frac K2$ and $a+b=2y-2x-\frac K2$.
Setting $K = 4kt$, we recover DisintegratingByParts's solution,
$$ u(x,t) = \frac{1}{\sqrt{4\pi t}}\int_{\mathbb R} \exp\left(\frac{-\big(y-x-\frac {4kt }2\big)^2}{4t}+kx+k^2t\right) dy = e^{kx+k^2t}.$$
By Widder's uniqueness result for non-negative solutions, this is the unique solution.
$u(x,t)=e^{kx+k^2t}$ looks like it might work.
In order to evaluate using the integral solution, write the terms in $y$ as $e^p$ where $p$ is a quadratic in $y$; then complete the square in $y$ and shift the integral to reduce to $\alpha y^2+\beta$ in the exponent. Then you should be able to see the solution.