First order ODE with Dirac delta funtcion

Question

I am looking for a direct method to solve this first order ODE with Dirac delta funtcion

$$\frac{dU(t)}{dt}+k^2U(t)=\frac{1}{\sqrt{2\pi}}\delta(t)$$

with the initial condition $U(0)=\frac{1}{\sqrt{2\pi}}$.

The solution to this problem is $$U(t)=\frac{1}{\sqrt{2\pi}}e^{-k^2t}$$

My try

The integrating factor for this ode is $$I=e^{\int k^2 dt }=e^{ k^2 t }$$

then multiplying both sides of the differential equation by $\,\,e^{ k^2 t }$, we get

$$\frac{d }{d t}\left(e^{ k^2 t } U(t)\right)=\frac{1}{\sqrt{2\pi}}e^{ k^2 t } \delta(t)$$ Integrating both sides, we have

$$e^{ k^2 t } U(t)=\frac{1}{\sqrt{2\pi}}\int{e^{ k^2 t } \delta(t)}dt+C$$

From here on, I am lost. Any suggestions?

Back ground of this problem

The above ode we got after applying Fourier transform to the following PDE $$u_{t}=u_{xx}+\delta{(x)}\delta{(t)}$$ with $u(x,0)=\delta(x)$.

Answer 1

This maybe an answer to my question but needs your conformation/validation.

Continuing from where I left $$e^{ k^2 t } U(t)=\frac{1}{\sqrt{2\pi}}\int{e^{ k^2 t } \delta(t)}dt+C$$ $$\Rightarrow U(t)=\bigg[\frac{1}{\sqrt{2\pi}}\int{e^{ k^2 t } \delta(t)}dt+C\bigg]e^{ -k^2 t }$$ Now recalling some properties of the Dirac delta function

1. $$\int_{-\infty}^{\infty}f(t)\delta(t)dt=f(0)$$
2. $$\int_{a}^{b}f(t)\delta(t)dt= \begin{cases} f(0) & \text{if } a<t<b\\ 0 & \text{if } x=0 \end{cases} $$ Now since we are dealing with an IVP, where $t\geq0$, then adopting the above properties "we can have" $$\int{e^{ k^2 t } \delta(t)}dt=e^{0}=1,$$

which gives $$U(t)=\bigg[\frac{1}{\sqrt{2\pi}}+C\bigg]e^{ -k^2 t }$$ Finally utilizing the initial condition $U(0)=\frac{1}{\sqrt{2\pi}}$, gives $C=0$, and hence $$U(t)=\frac{1}{\sqrt{2\pi}}e^{ -k^2 t }$$

Is this correct?