I tried to solve the following SDE but I'm struggling to utilize the fact that $X_1 = 0$.
Find the solution of the following differential equation: $$dX_t=\left(t+\frac{1}{t}\right)X_tdt + 2t\exp\left(\frac{t^2}{2}\right)dB_t$$ for $t \in (0,1]$ and $X_1 = 0$.
I used the ansatz: $$X_t = a(t)\left[x_0+\int_0^t b(t)dB_t\right]$$ and calculated the "solution": $$X_t = c_1t\exp\left(\frac{t^2}{2}\right)(x_0+\frac{2}{c_1}B_t)$$
Is this correct so far? I guess the assignment was faulty and it was meant to be $X_0=0$ such that the solution simplifies to: $$X_t=2t\exp\left(\frac{t^2}{2}\right)B_t$$ otherwise I don't know how to proceed with $X_1 = 0$.
Set $u(t)=\frac{t^2}2+\ln(t)$ then $u'(t)=t+\frac1t$ and $e^{u(t)}=t\exp(\frac{t^2}2)$ so that your differential equation reads $$ dX_t=u'(t)X_tdt+2e^{u(t)}dB_t\\ \\\iff\\ d(e^{-u(t)}X_t)=e^{-u(t)}(dX_t-u'X_t)=2dB_t $$ which directly integrates to $$ e^{-u(t)}X_t-e^{-u(1)}X_1=2(B_t-B_1) $$ and with $X_1=0$ $$ X_t=2t\exp(\frac{t^2}{2\,})(B_t-B_1) $$ which is not that far off from your expected general form of the solution. Note that $B_t-B_1$ is equal to $\tilde B_{t-1}$ for another Brownian motion.