By application of Feynman-Kac theorem $U$ has the representation
$$U(t,X_t)=e^{-r(T-t)}\mathbb{E}_{t}\left[\,\max\{X_T-K,0\}\,\right]$$
where $X_t$ satisfy the SDE
$$dX_t=\mu X_tdt+\sigma X_t dW_{t}^{\mathbb{P}}\tag 1$$
Now, we define a new measure $\mathbb{Q}$ by
$$d\mathbb{Q}=L_T\,d\mathbb{P}\quad$$
on $\mathcal{F}_T$ where
$$dL_t=\left(\frac{\mu-r}{\sigma}\right)L_t dW^{\mathbb{P}}_t.$$
By application of Girsanov theorem, we have
$$dW^{\mathbb{P}}_t=-\left(\frac{\mu-r}{\sigma}
\right)dt+dW^{\mathbb{Q}}_t\tag 2$$
$(1)$ and $(2)$
$$dX_t=r X_tdt+\sigma X_t dW_{t}^{\mathbb{Q}}.\tag 3$$
By application of Ito's lemma
$$\ln X_T=\ln X_t+\left( r-\frac{1}{2}\sigma ^{2} \right)(T-t)+\sigma (W_T-W_t)$$
Indeed we showed
$$\ln X_T\sim N\left(\ln X_t+\left( r-\frac{1}{2}\sigma ^{2} \right)(T-t)\,,\, \sigma^2(T-t)\right)\tag 4$$
therefore
$$Q(X_T<K)=Q(\ln X_T<\ln K)=N\left(\frac{\ln K-\ln X_t-\left( r-\frac{1}{2}\sigma ^{2} \right)(T-t)}{\sigma^2\sqrt{T-t}}\right)$$
we now $N(-x)=1-N(x)$, thus
$$Q({{X}_{T}}>K)=N\left(\frac{\ln \left(\frac{X_t}{K}\right)+\left( r-\frac{1}{2}\sigma ^{2} \right)(T-t)}{\sigma^2\sqrt{T-t}}\right)=N(d_2)\tag 5$$
Now we should change the measure $\mathbb{Q}$ to another measure $\mathbb{Q}^X$. Consider the Radon-Nikodym derivative
$$\frac{d\mathbb{Q}^X}{d\mathbb{Q}}=\frac{B_T/B_t}{X_T/X_t}$$
where
$$B_t=\exp\left(\int_{0}^{t}r\,du\right)=e^{rt}$$
as a result
$${{\mathbb{Q}}^{X}}({{X}_{T}}>K)=\int\limits_{K}^{+\infty }{d{{\mathbb{Q}}^{X}}}=\frac{{{e}^{-r(T-t)}}}{{{X}_{t}}}\int\limits_{K}^{+\infty }{{{X}_{T}}\,d\mathbb{Q}}=\frac{{{e}^{-r(T-t)}}}{{{X}_{t}}}\int\limits_{K}^{+\infty }{{{X}_{T}}{{f}_{{{X}_{T}}}}(x)dx} $$
we have
$$\mathbb{Q}^X(X_T>K)=\frac{e^{-r(T-t)}}{X_t}E^\mathbb{Q}[X_T|X_T>K]=N\left(\frac{\ln \left(\frac{X_t}{K}\right)+\left( r+\frac{1}{2}\sigma ^{2} \right)(T-t)}{\sigma^2\sqrt{T-t}}\right)$$
Indeed
$$\mathbb{Q}^X(X_T>K)=N(d_1)\tag 6$$
on the other hand
$$U(t,x)=e^{-r(T-t)}\mathbb{E}_{t}^{\mathbb{Q}}\left[\,\max\{X_T-K\},0\,\right]\tag 7$$
it is obvious
$$\max\{X_T-K,0\}=(X_T-K)\mathbb{1}_{\{X_T>K\}}$$
then
$$U(t,X_t)=e^{-r(T-t)}\mathbb{E}_{t}^{\mathbb{Q}}\left[X_T\mathbb{1}_{\{X_T>K\}}\right]-e^{-r(T-t)}\mathbb{E}_{t}^{\mathbb{Q}}\left[K\mathbb{1}_{\{X_T>K\}}\right]$$
as a result
$$U(t,X_t)=X_t\,\mathbb{E}_{t}^{\mathbb{Q}}\left[\frac{X_T/X_t}{B_T/B_t}\mathbb{1}_{\{X_T>K\}}\right]-Ke^{-r(T-t)}\mathbb{E}_{t}^{\mathbb{Q}}\left[\mathbb{1}_{\{X_T>K\}}\right]$$
in other words
$$U(t,X_t)=X_t\mathbb{E}_{t}^{\mathbb{Q}^X}\left[\mathbb{1}_{\{X_T>K\}}\right]-Ke^{-r(T-t)}\mathbb{E}_{t}^{\mathbb{Q}}\left[\mathbb{1}_{\{X_T>K\}}\right]$$
so
$$U(t,X_t)=X_t\mathbb{Q}^X(X_T>K)-Ke^{-r(T-t)}\mathbb{Q}(X_T>K)\tag 8$$
$(5)$ ,$(6)$ and $(8)$
$$U(t,X_t)=X_tN(d_1)-Ke^{-r(T-t)}N(d_2)$$
