Let $B$ be the brownian motion. How can I show that $dX_{t} = \sqrt{2c\lambda} dB_{t}-\lambda X_{t}dt$ is solved by
$$ X_{t}=X_{0}e^{-\lambda t}+\sqrt{2c\lambda}\int\limits_{0}^{t}e^{-\lambda(t-s)}dB_{s}$$
My attempt:
I wanted to use Ito's formula, more particularly, the integration by parts formula:
$$e^{\lambda t}X_{t}=X_{0}+\int\limits_{0}^{t}e^{\lambda s}dX_{s}+\int\limits_{0}^{t}X_{s}d(e^{\lambda s})=X_{0}+\int\limits_{0}^{t}e^{\lambda s}dX_{s}+\int\limits_{0}^{t}X_{s}\lambda e^{\lambda s}ds\\=X_{0}+\int\limits_{0}^{t}e^{\lambda s}(X_{0}e^{-\lambda s}+\sqrt{2c\lambda}\int\limits_{0}^{s}e ^{-\lambda(s-u)}dB_{u})ds+\int\limits_{0}^{t}X_{s}\lambda e^{\lambda s}ds \\= X_{0}+\int\limits_{0}^{t}(X_{0}+\sqrt{2c\lambda}\int\limits_{0}^{s}e^{\lambda u}dB_{u})ds+\int\limits_{0}^{t}X_{s}\lambda e^{\lambda s}ds$$
then multiplying by $e^{-\lambda t}$, we obtain:
$$X_{t}=e^{-\lambda t}X_{0}+\int\limits_{0}^{t}(e ^{-\lambda t}X_{0}+\sqrt{2c\lambda}\int\limits_{0}^{s}e^{-\lambda(t-u)}dB_{u})ds+\int\limits_{0}^{t}X_{s}\lambda e^{\lambda s}ds$$
It feels like I am on the wrong track but I do not know what I am doing wrong.
If you enter the expression for $dX_s$ into $$ X_0+\int_0^te^{\lambda s}\,dX_s+\int_0^t X_s\lambda e^{\lambda s}\,ds $$ you get \begin{align} e^{\lambda t}X_t&=X_0+\int_0^te^{\lambda s}\sqrt{2c\lambda}\,dB_s-\int_0^te^{\lambda s}\lambda X_s\,ds+\int_0^tX_s\lambda e^{\lambda s}\,ds\\ &=X_0+\int_0^te^{\lambda s}\sqrt{2c\lambda}\,dB_s \end{align} which immediately leads to the solution.
Alternatively, you can apply Ito's formula to the proposed solution \begin{align} X_t&=X_0e^{-\lambda t}+\sqrt{2c\lambda}\int_0^te^{-\lambda(t-s)}\,dB_s\\ &=e^{-\lambda t}\left\{X_0+\sqrt{2c\lambda}\int_0^te^{\lambda s}\,dB_s\right\} \end{align} which gives \begin{align} dX_t&=-\lambda e^{-\lambda t}\left\{X_0+\sqrt{2c\lambda}\int_0^te^{\lambda s}\,dB_s\right\}\,dt+e^{-\lambda t}\sqrt{2c\lambda}e^{\lambda t}\,dB_t\\ &=-\lambda X_t\,dt+\sqrt{2c\lambda}\,dB_t\,. \end{align}