Use Ito's formula to determine the stochastic differential equation satisfied by $V_t$ A stochastic process $V_t$ is defined by
$$V_t  =\sqrt{t(t+W_t^2)}$$
$W_t$ is the Wiener process and $t$ denotes the time ($t > 0$).
Use Ito's formula to determine the stochastic differential equation satisfied by $V_t$.
I have taken the known process to be just $W$. Then $a(x,t) = 0$ and $b(x,t)=1$.
The process $V$ is: 
$$V_t = h(W_t,t)\ \text{where}\ h(x,t) = \sqrt{t(t+x^2)}$$
I then calculated the partial derivatives of $h(x,t)$ and substituted all my values into Ito’s formula.
However, no matter how many times I try, I don’t seem to get the correct answer and I am unsure as to why.
 A: Set $$f(t,x) := \sqrt{t \cdot (t+x^2)}. \DeclareMathOperator \sgn{sgn}$$
By the chain rule,
$$\begin{align*} \frac{\partial}{\partial x} f(t,x) &= \frac{xt}{f(t,x)} \tag{1} \\
\frac{\partial^2}{\partial x^2} f(t,x) &= \frac{t}{f(t,x)} - \frac{x^2 t^2}{f(t,x)^3}\tag{2} \\
\frac{\partial}{\partial t} f(t,x) &= \frac{2t+x^2}{2f(t,x)} \tag{3}\end{align*}$$
In order to find the corresponding SDE, we have to express the expressions on the right-hand side in terms of $f(t,x)$ and $t$. From the definition of $f$, it follows that
$$x^2 = \frac{f(t,x)^2}{t}-t \tag{4}$$
If we plug $(4)$ into $(2)$ and $(3)$, we get
$$\begin{align*}
\frac{\partial^2}{\partial x^2} f(t,x) &= \frac{t}{f(t,x)} - \left( \frac{f(t,x)^2}{t}-t \right) \frac{t^2}{f(t,x)^3} = \frac{t^3}{f(t,x)^3} \\
\frac{\partial}{\partial t} f(t,x) &= \frac{t}{f(t,x)} + \left( \frac{f(t,x)^2}{t}-t \right) \frac{1}{2 f(t,x)} = \frac{1}{2} \frac{t}{f(t,x)} + \frac{f(t,x)}{t}. \end{align*}$$
Similarly,
$$\begin{align*} \frac{\partial}{\partial x} f(t,x) &= \frac{t}{f(t,x)} \sgn(x) \sqrt{\frac{f(t,x)^2}{t}-t} = \sgn(x) \sqrt{t- \frac{t^3}{f(t,x)^2}}. \end{align*}$$
Now if we apply Itô's formula, i.e.
$$V_t = f(t,W_t) = \int_0^t \frac{\partial}{\partial x} f(s,W_s) \, dW_s + \int_0^t \left( \frac{1}{2} \frac{\partial^2}{\partial x^2} f(s,W_s) + \frac{\partial}{\partial t} f(s,W_s) \right) \, ds,$$
we find that
$$V_t = \int_0^t \sgn(W_s) \sqrt{s} \sqrt{1- \frac{s^2}{V(s)^2}} \, dW_s + \int_0^t \left( \frac{V(s)}{s} + \frac{1}{2} \frac{s}{V(s)}+\frac{1}{2} \frac{s^3}{V(s)^3} \right) \, ds.$$
Note that the diffusion coefficient depends on the Brownian motion $(W_t)_{t \geq 0}$. Since $V_s$ does not contain any information whether $W_s$ is positive or negative, we cannot expect to express $\sgn(W_s)$ in terms of $V_s$. This means that we cannot write $(V_t)_{t \geq 0}$ as the solution of a stochastic differential equation.
A: I took a slightly different route
$$S_t = V_t^2 = t(t+x^2)$$
$$x^2 = \frac{V_t^2}{t} - t$$
$$dS_t = 2 V_t dV_t=\frac{\partial S_t}{\partial t} dt + \frac{\partial S_t}{\partial x} dW_t + \frac{1}{2} \frac{\partial^2 S_t}{\partial x^2} dt$$
$$\frac{\partial S_t}{\partial t} = 2t + x^2$$
$$\frac{\partial S_t}{\partial x} = 2tx$$
$$\frac{\partial^2 S_t}{\partial x^2} = 2t$$
and so
$$\begin{align*}
2 V_t dV_t &= \left(t + \frac{V_t^2}{t}\right)dt + 2t\left(\frac{V_t^2}{t} - t\right)^\frac{1}{2} dW_t + tdt\\
&= \left(2t + \frac{V_t^2}{t}\right)dt + 2\left(tV_t^2 - t^3\right)^\frac{1}{2} dW_t
\end{align*}$$
$$dV_t = \left(\frac{t}{V_t} + \frac{V_t}{2t}\right)dt + \left(t-\frac{t^3}{V_t^2}\right)^\frac{1}{2}dW_t$$
I'm slightly disconcerted that my $dt$ coefficient looks so different to saz's. I'll need to do some rearranging to check whether or not they actually agree...
Edit: As pointed out in the comments I missed the implicit absolute value in the square root and left out the second order term in $dS_t$. Blast.
