Let $(B_t)_{t \in [0,T]}$ be standard brownian motion, and let $(Y_t)_{t \in [0,T]}$ be a stochastic process in $(\Omega, \mathscr F, \mathbb P)$.
Without using the general linear SDE formula, solve the SDE:
$$dY_t = tY_tdt + e^{t^2/2}dB_t$$ by considering $$Y_t = a(t) \left(Y_0 + \int_0^t b(s) dB_s \right)$$
with initial condition $Y_0 = \alpha \in \mathbb R$, where $a(t)$ and $b(t)$ are differential functions that are not random.
Hint: Consider
$$Y_t = a(t) \left(Y_0 +\int_{0}^{t} b(s) dB_s \right)$$
What I tried for solving:
$$dY_t = a'(t)Y_0 dt + a'(t) \ dt \ b(t) \ dB_t + a(t)b(t)dB_t + a'(t) dt \int_0^t b(s) dB_s$$
$$= a'(t)Y_0dt + a(t)b(t)dB_t + a'(t) dt \int_0^t b(s) dB_s$$
$$= a'(t) \alpha dt + a(t)b(t)dB_t + a'(t) dt \int_0^t b(s) dB_s$$
$$= Y_t\frac{a'(t)}{a(t)}dt + a(t)b(t)dB_t$$
Equating
$$Y_t\frac{a'(t)}{a(t)}dt + a(t)b(t)dB_t = tY_tdt + e^{t^2/2}dB_t,$$
we get that
$$a(t) = Ae^{t^2/2}$$
where $A = \pm e^C$ where $C \in \mathbb R$ and
$$b(t) = 1/A$$
Hence,
$$Y_t = \alpha Ae^{t^2/2} + e^{t^2/2}B_t$$
Is that right? I find that A kind of weird. I think it's supposed to cancel out.
What I tried for proving uniqueness:
- Show that $E[Y_0^2] < \infty$:
$$E[Y_0^2] = E[\alpha^2] = \alpha^2 < \infty$$
- Show $\exists K \in \mathbb R$ s.t.
$|tx| + |e^{t^2/2}| \leq K(|x|+1)$
$|tx-ty| \leq K(|x-y|)$:
We have $$|t| = t \le T \to e^{t^2/2} \le e^{T^2/2}$$
Choose $$K = e^{T^2/2}$$
Is that right?
Note that
$$Y_t = \alpha A e^{t^2/2} + e^{t^2/2} B_t$$
has to satisfy the initial condition $Y_0 = \alpha$. Thus,
$$Y_0 = \alpha A \stackrel{!}{=} \alpha,$$
i.e. $A=1$. Consequently,
$$Y_t = e^{t^2/2} (\alpha+B_t).$$
If you want to check whether this is indeed a solution to the given SDE, then just apply Itô's formula with
$$f(t,x) := e^{t^2/2} (\alpha+x),$$
i.e.
$$Y_t -Y_0 = \int_0^t e^{s^2/2} \, dB_s + \int_0^t s e^{s^2/2} (\alpha+B_s) \, ds = \int_0^t e^{s^2/2} \, dB_s + \int_0^t s Y_s \, ds.$$
This shows that $(Y_t)_{t \geq 0}$ solves the given SDE.
(And, yeah, the constants for the proof of the uniqueness are okay.)