Solving a stochastic differential equation using Ito’s formula

Question

I was wondering wether anyone could help me solve this question.

Let $B=(B_t)t\geq0$ be a standard Brownian motion started at zero. Consider the stochastic differential equation:

$dX_t = (1+ 3X_t)dt + (1+5X_t)dB_t$,

for a stochastic process $X=(X_t)_{t\geq0}$ where $X_0=1$.

Verify by Ito’s formula that this solution is given by

$X_t = Y_t(1-4\int_0^t\frac{1}{Y_s}ds+ \int_0^t\frac{1}{Y_s}dB_s)$ for $t\geq0$, where the process $Y=(Y_t)_{t\geq0}$ solves

$dY_t = 3Y_tdt + 5Y_tdB_t$, with $Y_0=1$

I’m not sure how to apply Ito’s formula to obtain the solution the question is after? Any help would be great!

Answer 1

Let $A_t := 1 - 4\int_0^t \frac 1{Y_s}ds + \int_0^t \frac 1{Y_s}dB_s$, so $dA_t = -4 \frac{1}{Y_t}dt + \frac 1{Y_t}dB_t.$ Note that the equality for $dA_t$ is just notation for how we defined $A_t$. Then Ito's product formula applied to $X_t = Y_t A_t$ gives

\begin{align*} dX_t &= Y_t dA_t + A_t dY_t + dA_tdY_t \\ &= (-4dt + dB_t) + (3Y_tA_t dt + 5Y_tA_t dB_t) + 5dt \\ &= (1+3Y_tA_t)dt + (1+5Y_tA_t)dB_t \\ &= (1+3X_t)dt + (1+5X_t)dB_t. \end{align*}