Strong solution of stochastic differential equation Consider the stochastic differenctial equation:
$dX_t=\frac34 X_t^2 dt-X_t^{3/2}dW_t$.
How to find a strong solution?
 A: If we want to solve an SDE of the form
$$dX_t = \sigma(X_t) \, dW_t + b(X_t) \, dt$$
i.e. an SDE which does not depend explicitly on the time $t$, the transformation
$$f(x) := \int^x \frac{1}{\sigma(y)} \, dy$$
is always worth a try. Here, $\sigma(y)=-y^{3/2}$; hence we consider
$$f(x) =-\int_x^\infty \frac{1}{\sigma(y)} \, dy =\int_x^\infty y^{-3/2} \, dy = 2 \frac{1}{\sqrt{x}}.$$
Then,
$$f'(x) = -\frac{1}{x^{3/2}} \qquad f''(x) = \frac{3}{2} \frac{1}{x^{5/2}}.$$
By Itô's formula,
$$\begin{align*} f(X_t)-f(X_0) &= -\int_0^t \frac{1}{X_s^{3/2}} \, dX_s + \frac{3}{4} \int_0^t \frac{1}{X_s^{5/2}} \, d\langle X \rangle_s \\ &= \int_0^t dW_s - \frac{3}{4} \int_0^t X_s^{1/2} \, ds + \frac{3}{4} \int_0^t X_s^{\frac{1}{2}} \, ds \\ &= W_t \end{align*}$$
where we used that the quadratic variation $\langle X \rangle_s$ has infinitesimal variation $X_s^3 \, ds$. This shows that
$$X_t = f^{-1} \bigg( W_t+ f(X_0) \bigg) = \frac{4}{(W_t+f(X_0))^2} = \frac{X_0}{(1+\frac12\sqrt{X_0}\cdot W_t)^2}$$
for every $t\lt T$, where
$$
T=\inf\{t\gt0\mid W_t=-2/\sqrt{X_0}\}.
$$
