# Solve SDE $dX_t = (e^{-\gamma t} - \gamma X_t) dt + 2 e^{-\gamma t/2} \sqrt{X_t} dW_t$

Solve the SDE given by: $dX_t = (e^{-\gamma t} - \gamma X_t) dt + 2 e^{-\gamma t/2} \sqrt{X_t} dW_t$.

My attempt

Following the hint of my professor: suppose $X_t = e^{\gamma t} g(W_t)$. Then we can write: $$dX_t = (e^{-\gamma t}((1/2)g''(W_t))-\gamma X_t) dt + (e^{-\gamma t} g'(W_t)) dW_t$$

By identification,

$$1=\dfrac{g''(W_t)}{2}$$ and, $$g'(W_t) = 2 (g(W_t))^{1/2}$$ with initial condition $$g(0) = 1$$

EDIT If we consider the equations above as "non-stochastic" DE's, we can attempt to find the solution to the system. The solution to the first equation is a simple polynomial, but the solution to the second equation (the non-linear ODE) is very long. How can I proceed?

The first equation tells us that $g$ is a polynomial of (at most) second order, i.e. we can make the ansatz

$$g(x) = a+bx+cx^2.$$

Since $g(0) \stackrel{!}{=} 1$, we have $a=1$. Moreover, as

$$g'(x) = b+2cx,$$

the condition

$$g'(x)^2 = 4 g(x) \tag{1}$$

is equivalent to

$$b^2+4bcx + 4c^2 x^2 = 4 + 4bx + 4cx^2.$$

So,

$$c^2 \stackrel{!}{=} c \qquad b \stackrel{!}{=} bc \qquad b^2 \stackrel{!}{=} 4.$$

This implies $c=1$ and $b = \pm 2$. In $(1)$, we have lost the information on the sign of $b$; unfortunately, neither $$g_1(x) := 1+2x+x^2 = (x+1)^2$$ nor $$g_2(x) = 1-2x+x^2 = (x-1)^2$$ satisfies the condition

$$g'(x) = 2 \sqrt{g(x)}$$

(as this would imply in particular $g'(x) \geq 0$, which is - in both cases- not satisfied for $x <-1$). This means that we don't get any solution using this approach.