I am trying to solve: $ dY_t = dt + 2 \sqrt{Y_t} dW_t$ with $Y_0 = y_0 > 0$, where $W$ is one dimentional standard Brownian motion. Using the clue in this linked question, I use the substitution, $Z_t=\exp(-\sqrt{Y_t})$. I derive the derivatives as follows.
\begin{align*} f(y)&=\exp(-\sqrt{y}),\\ f'(y)&=\frac{\exp(-\sqrt{y})}{-2\sqrt{y}},\\ f''(y) &= \frac{-2\sqrt{y}\frac{\exp(-\sqrt{y})}{-2\sqrt{y}}-\exp(-\sqrt{y})\frac{-1}{\sqrt{y}}}{4y},\\ &=\exp(-\sqrt{y}) \frac{1+\frac{1}{\sqrt{y}}}{4y}. \end{align*}
Using Ito's formula, I get: \begin{align*} dZ_t &= f'(Y_t)dY_t+\frac12f''(Y_t)4Y_t dt,\\ dZ_t &= \frac{-Z_t}{2 \sqrt{Y_t}}dt -Z_tdW_t+\frac12Z_t\left(1+\frac{1}{\sqrt{Y_t}}\right)dt\\ &= \frac12Z_t dt-Z_tdW_t. \end{align*}
The solution is $Z_t = Z_0 \exp(-W_t)$, or $Y_t=\left(\sqrt{Y_o}+W_t\right)^2.$, or using $X_t=\sqrt{Y_t}$, $X_t-X_0=W_t$.
Is the above solution correct? I am confused because the last equation corresponds to $dX_t=dW_t$, which is different from the equation $dX_t=\frac{1}{2X_t}dt+dW_t$ - do they both give rise to the same SDE?