Show that $X_{t} = \sqrt{c}e^{-\lambda t}\beta_{e^{2\lambda t}}$ solves $dX_{t}=\sqrt{2c\lambda}dB_{t}-\lambda X_{t}dt$

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 $X_t$.