Solution of two (first) SDEs. I'm about to study SDE's for the first time and I'm kinda having troubles "guessing"/"finding" solutions. Also I don't really know how and when analogies to simple ODEs are allowed (e.g. to get a first glimps how the solution could look like, later testing with Ito).
I got two SDE's that bother me atm:
$\mathbb{d}X_t=|X_t|\mathbb{d}B_t+\frac{X_t}{2}\mathbb{d}t, X_0=0$ and
$\mathbb{d}X_t=\sqrt{1+X_t^2}\mathbb{d}B_t+\frac{X_t}{2}\mathbb{d}t, X_0=x$
I tried to get rid of the "dt"-parts with the common ODE trick (multiply with exp(..)), with no effect. Guess the $dB_t$ part should be linear in $X_t$ for that trick to work?
I would very much appreciate any tips and/or solutions, in particular some intuition how to e.g. make educated guesses, use analogies from ODEs,...
 A: Let's start with the SDE
$$dX_t = |X_t| \, dB_t+ \frac{X_t}{2} \, dt \qquad X_0=0$$
This one is pretty easy: As the coefficients are globally Lipschitz continuous, there exists a unique solution. Obviously, $X_t := 0$ is a solution, and so we are done.
Now consider the more interesting SDE
$$dX_t = \sqrt{1+X_t^2} \, dB_t + \frac{X_t}{2} \, dt, \qquad X_0=x \tag{2}$$
Without loss of generality, we may assume $x=0$. First of all, we note that the SDE is of the form
$$dX_t = \sigma(X_t) \, dB_t + b(X_t) \, dt \tag{3}$$
i.e. the coefficients do not depend (explicitely) on the time $t$. For this type of SDEs there exist conditions for a transformation into a linear SDEs - and these can be solved explicitely. Since you are interested in the intuition, I will try to motivate the Ansatz and state the corresponding result below:
Clearly, it would be nice to simplify the stochastic-integral term. Itô's formula shows that for $$Z_t := f(t) \cdot \int_0^{X_t} \frac{1}{\sigma(y)} \, dy$$ we get $$Z_t-Z_0 = \int f(s) \, dB_s + \int_0^t \ldots \, ds,$$
so (at least) the stochastic integral is fine. It remains to make the $ds$-term as nice as possible. Obviously, this term depends on our choice of $f$.The function $f(t)=e^{ct}$ might be a good choice (as it "reproduces" itself if we differentiate with respect to $t$). Indeed, there is the following (general) result:

Suppose that the coefficients $b$, $\sigma$ satisfy $$0 = \frac{d}{dx} \left( \sigma(x) \cdot \left( - \frac{d}{dx} \frac{b(x)}{\sigma(x)} + \frac{1}{2} \sigma''(x) \right) \right).$$ Then the transformation $$Z_t = e^{c t} \int_0^{X_t} \frac{1}{\sigma(y)} \, dy\tag{4}$$ converts $(3)$ into a linear SDE (for $(Z_t)_t$).

(see René L. Schilling/Lothar Partzsch: Brownian motion-An Introduction to Stochastic Processes, Example 18.7)
It is not difficult to check that for the SDE $(2)$, the condition is satisfied. Moreover, some straight-forward calculations show that $c=0$ is the best choice and we obtain
$$Z_t = \int_0^t \, dB_s = B_t$$
By definition,
$$B_t = Z_t \stackrel{(4)}{=} \int_0^{X_t} \frac{1}{\sqrt{1+y^2}} \, dy.$$
As $y \mapsto \frac{1}{\sqrt{1+y^2}}$ is the derivative of $y \mapsto \text{arsinh} \, y$, we get
$$B_t = \text{arsinh}(X_t)$$
i.e.
$$X_t = \sinh(B_t)$$
Let me finally remark that there are different approaches but at least the ones I know do not work out for the given SDE.
A: The Ito formula requires that, when you create a Taylor series for $X_t=X(t, W_t)$ you have not only the partials w.r.t. $t$ and $W_t$, but the second partial $\frac {\partial ^2X(t, W_t)}{\partial X_t^2}$ that is not of lesser order than the first derivatives.  
Unless you have a very simple linear mapping, such that $\frac {\partial ^2X(t, W_t)}{\partial X_t^2}$ is zero, the solutions are not the same as for regular differential equations.  They often look about the same but with an 'Ito correction term' that comes from the second partial derivative.
It will be the same, for example, if you have $dX_t=\alpha t+\beta dW_t$ since $\frac {\partial ^2X(t, W_t)}{\partial X_t^2}$ is zero.  But if it isn't, the normal solutions by inspection don't work.  
