\[ -a y(x)^3-\frac {b}{x^{3/2}}+y'(x)=0 \] ✓ Mathematica : cpu = 0.0994194 (sec), leaf count = 99
\[\text {Solve}\left [-2 \text {RootSum}\left [-2 \text {$\#$1}^3+\text {$\#$1} \sqrt [3]{-\frac {1}{a b^2}}-2\& ,\frac {\log \left (y(x) \sqrt [3]{\frac {a x^{3/2}}{b}}-\text {$\#$1}\right )}{\sqrt [3]{-\frac {1}{a b^2}}-6 \text {$\#$1}^2}\& \right ]=\frac {a x \log (x)}{\left (\frac {a x^{3/2}}{b}\right )^{2/3}}+c_1,y(x)\right ]\]
✗ Maple : cpu = 0. (sec), leaf count = 0 , could not solve
dsolve(diff(y(x),x)-a*y(x)^3-b*x^(3/2) = 0,y(x))
\begin {equation} y^{\prime }\left ( x\right ) =ay^{3}+bx^{-\frac {3}{2}}\tag {1} \end {equation}
This can be transformed to Abel first order non-linear ode as follows. Let \(y\left ( x\right ) =x^{-\frac {1}{2}}\eta \left ( \xi \right ) \) where \(\xi =\ln x\) hence
\begin {align*} \frac {dy}{dx} & =-\frac {1}{2}x^{-\frac {3}{2}}\eta \left ( \xi \right ) +x^{-\frac {1}{2}}\frac {d\eta }{d\xi }\frac {d\xi }{dx}\\ & =-\frac {1}{2}x^{-\frac {3}{2}}\eta \left ( \xi \right ) +x^{-\frac {1}{2}}\frac {d\eta }{d\xi }\frac {1}{x}\\ & =-\frac {1}{2}x^{-\frac {3}{2}}\eta \left ( \xi \right ) +x^{-\frac {3}{2}}\frac {d\eta }{d\xi } \end {align*}
Substituting in (1) gives
\begin {align*} -\frac {1}{2}x^{-\frac {3}{2}}\eta \left ( \xi \right ) +x^{-\frac {3}{2}}\frac {d\eta }{d\xi } & =a\left ( x^{-\frac {1}{2}}\eta \left ( \xi \right ) \right ) ^{3}+bx^{-\frac {3}{2}}\\ -\frac {1}{2}x^{-\frac {3}{2}}\eta \left ( \xi \right ) +x^{-\frac {3}{2}}\frac {d\eta }{d\xi } & =ax^{-\frac {3}{2}}\eta ^{3}\left ( \xi \right ) +bx^{-\frac {3}{2}}\\ -\frac {1}{2}\eta +\eta ^{\prime } & =a\eta ^{3}+b\\ \eta ^{\prime } & =b+\frac {1}{2}\eta +a\eta ^{3} \end {align*}
This is Abel first kind. In general form it is
\[ \eta ^{\prime }=f_{0}+f_{1}\eta +f_{2}\eta ^{2}+f_{3}\eta ^{3}\]
Where in this case \(f_{0}=b,f_{1}=\frac {1}{2},f_{2}=0,f_{3}=a\). Using Maple, the solution to the above is (I need to learn how to solve Able by hand more) is implicit, given as
\[ \eta =\xi -\int ^{\eta \left ( \xi \right ) }\frac {1}{b+\frac {1}{2}z+az^{3}}dz+C \]
Where \(C\) is constant of integration. Hence, since \(y\left ( x\right ) =x^{-\frac {1}{2}}\eta \left ( \xi \right ) \), then \(\eta \left ( \xi \right ) =\sqrt {x}y\) and the above becomes
\begin {align*} \sqrt {x}y & =\ln x-\int ^{\sqrt {x}y}\frac {1}{b+\frac {1}{2}z+az^{3}}dz+C\\ y\left ( x\right ) & =\left ( \ln x-\int ^{\sqrt {x}y}\frac {1}{b+\frac {1}{2}z+az^{3}}dz+C\right ) \frac {1}{\sqrt {x}} \end {align*}
DId not verify. Need to look more into this later.