Your solution seems fine. What does the rest of the error message say?
[tex]\displaystyle y^{1/2}\frac{\mathrm dy}{\mathrm dx} + y^{3/2} = 1[/tex]
Substitute
[tex]z(x)=y(x)^{3/2} \implies \dfrac{\mathrm dz}{\mathrm dx}=\dfrac32y(x)^{1/2}\dfrac{\mathrm dy}{\mathrm dx}[/tex]
to transform the ODE to a linear one in z :
[tex]\displaystyle \frac23\frac{\mathrm dz}{\mathrm dx} + z = 1[/tex]
Divide both sides by 2/3 :
[tex]\displaystyle \frac{\mathrm dz}{\mathrm dx} + \frac32z = \frac32[/tex]
Multiply both sides by the integrating factor, [tex]e^{3x/2}[/tex] :
[tex]\displaystyle e^{3x/2}\frac{\mathrm dz}{\mathrm dx} + \frac32 e^{3x/2}z = \frac32 e^{3x/2}[/tex]
Condense the left side into the derivative of a product :
[tex]\displaystyle \frac{\mathrm d}{\mathrm dx}\left[e^{3x/2}z\right] = \frac32 e^{3x/2}[/tex]
Integrate both sides and solve for z :
[tex]\displaystyle e^{3x/2}z = \frac32 \int e^{3x/2}\,\mathrm dx \\\\ e^{3x/2}z = e^{3x/2} + C \\\\ z = 1 + Ce^{-3x/2}[/tex]
Solve in terms of y :
[tex]y^{3/2} = 1 + Ce^{-3x/2}[/tex]
Given that y (0) = 16, we have
[tex]16^{3/2} = 1 + Ce^0 \implies C = 16^{3/2}-1 = 63[/tex]
so that the particular solution is
[tex]\boxed{y^{3/2} = 1 + 63e^{-3x/2}}[/tex]
