I am trying to solve the following SDE: $dX_{t} = cX_{t}dt + \sqrt{a + bX_{t}^2}dW_{t}$, where a and b are positive constants, $W_t$ is a Wiener process, $X_0$ is given.
I tried the trick to multiply $e^{-ct}$ on both sides and got $e^{-cT}X_{T} - X_0 = e^{-cT}\sqrt{a+bX_{T}^2}W_T$. Is this the only way of solving this SDE? Is there any other way of solving it where I can get a solution in the form $X_{t} = f(X_{0}, W_{t})$, i.e. RHS only depends on $X_{0}$ and $W_{t}$?
First we "remove" the drift as done in 5.5.B Shreve-Karatzas. We have the scale function to be
$$p''(x)=\frac{cx}{\sqrt{a+bx^{2}}}=p'(x),p(0)=0\Rightarrow p(x)=c_{1}\int_{1}^{x}exp\left(-2c\sqrt{a+bs^{2}}/b \right)ds,$$
and so by 5.13 prop, the process $Y_{t}:=p(X_{t})$ satisfies
$$Y_{t}=Y_{0}+\int_{0}^{t}\tilde{\sigma}(Y_{s})dW_{s},$$
for $\tilde{\sigma}(y):=p'(q(y))\sigma(q(y))$ and $q(y)$ the inverse of $p$.
And as mentioned in 5.21 Remark, the process $Y$ is just a time-changed Brownian motion (by the Dubins-Schwartz theorem). So $X_{t}=q(B_{g_{t}})$ for $g(t):=\int_{0}^{t}\tilde{\sigma}^{2}(s)ds$.