${y}^{\prime}-xy=x{y}^{3/2}\phantom{\rule{thinmathspace}{0ex}},y(1)=4$

$\begin{array}{rlr}{y}^{\prime}-xy=& x{y}^{3/2}& \\ {\displaystyle \frac{dy}{dx}}=& x(y+{y}^{3/2})& \\ {\displaystyle \frac{dy}{(y+{y}^{3/2})}}=& x\phantom{\rule{thinmathspace}{0ex}}dx& \\ -2\mathrm{ln}{\displaystyle \frac{1+\sqrt{y}}{\sqrt{y}}}=& {\displaystyle \frac{{x}^{2}}{2}}+c& \\ y=& {\displaystyle \frac{-1}{(1-{\mathrm{e}}^{{\textstyle \frac{-{x}^{2}}{4}}+c}{)}^{2}}}& \\ 4=& {\displaystyle \frac{-1}{(1-{\mathrm{e}}^{{\textstyle \frac{-1}{4}}+c}{)}^{2}}}& \\ c=& -0.655& \\ y=& {\displaystyle \frac{-1}{(1-{\mathrm{e}}^{{\textstyle \frac{-{x}^{2}}{4}}-0.655}{)}^{2}}}& \end{array}$

This is what I'm getting in Matlab

$\left(\begin{array}{c}\frac{{(\text{tanh}(\frac{{x}^{2}}{8}+\text{atanh}\left(3\right)-\frac{1}{8})+1)}^{2}}{4}\\ \frac{{(\text{tanh}(-\frac{{x}^{2}}{8}+\text{atanh}\left(5\right)+\frac{1}{8})-1)}^{2}}{4}\end{array}\right)$