Show that $X_{t} = \sqrt{c}e^{-\lambda t}\beta_{e^{2\lambda t}}$ solves $dX_{t}=\sqrt{2c\lambda}dB_{t}-\lambda X_{t}dt$ Let $(\beta_{t})$ and $(B_{t})$ be standard Brownian motions. How can I show that show that $X_{t} = \sqrt{c}e^{-\lambda t}\beta_{e^{2\lambda t}}$ solves $dX_{t}=\sqrt{2c\lambda}dB_{t}-\lambda X_{t}dt$?
My attempt:
$$ dX_{t}=d\left(\sqrt{c}e^{-\lambda t}\beta_{e^{2\lambda t}}\right)=\sqrt{c}d(e^{-\lambda t}\beta_{e^{2\lambda t}})=\sqrt{c}\left( e^{-\lambda t}d(\beta_{e^{2\lambda t}})+\beta_{e^{2\lambda t}}d(e^{-\lambda t})\right)$$
Now clearly $d(e^{-\lambda t})=-\lambda e^{-\lambda t }dt$. But I do not know how to compute $d(\beta_{e^{2 \lambda t}})$.
Personally, I want to think of $ \beta_{e^{2 \lambda t}}$ and $\beta_{g(t)}=f(t,\beta_{t})$ which is $C^{2}$, but it seems very suspicious.
 A: I am going to denote $Y_t = \sqrt{c} e^{-\lambda t} \beta_{e^{2\lambda t}} $ and show that $Y_t$ is a weak solution of the SDE for $X_t$. Notice that $Y_0 = \sqrt{c} \beta_1 \sim N(0, c)$, so we may as well assume that $X_0 \sim N(0, c)$ as well, independent of $(B_t)$.
Consider the function $f(x,t) = xe^{\lambda t}$ and set $Z_t = f(X_t, t)$. By Itô's lemma we get;
$$\begin{align*}
dZ_t &= \lambda X_t e^{\lambda t} dt + e^{\lambda t} dX_t = \sqrt{2 c \lambda } e^{\lambda t} dB_t
\end{align*}$$
Thus, $$Z_t = Z_0 + \sqrt{2c\lambda } \int_0^t e^{\lambda r}dB_r $$ After substituting $Z_t = X_t e^{-\lambda t}$, this implies that $$X_t = e^{\lambda t} X_0 + \sqrt{2c \lambda }e^{-\lambda t} \int_0^t e^{\lambda r} dB_r$$
Recall that Itô integrals of $L^2$ deterministic functions are Gaussian processes with mean zero, so that $X_t$ is a Gaussian process with mean zero and covariance
$$\begin{align*}
E(X_t X_s) &= e^{-\lambda t} e^{-\lambda s} E(X_0^2) + 2c\lambda e^{-\lambda t} e^{-\lambda s} \int_0^{\min (t,s)} e^{2\lambda r} dr \\
&= e^{-\lambda t} e^{-\lambda s} c + ce^{-\lambda t} e^{-\lambda s} (e^{2\lambda \min (t,s)} - 1) \\
&= c e^{-\lambda t}e^{-\lambda s}e^{2\lambda \min (t,s)}
\end{align*}$$
Notice that $Y_t$ is also a Gaussian process with mean zero and covariance $$E(Y_t Y_s) = c e^{-\lambda t} e^{-\lambda s} \min (e^{2\lambda t}, e^{2\lambda s})=c e^{-\lambda t} e^{-\lambda s} e^{2\lambda \min (t,s)} = E(X_tX_s)$$
Since the distribution Gaussian processes is determined by its first two moments, $Y_t$ and $X_t$ agree in distribution, showing that $Y_t$ is a weak solution of the SDE for $X_t$.
