\[ a y(x)^2+b x^2+x y'(x)-y(x)=0 \] ✓ Mathematica : cpu = 0.113448 (sec), leaf count = 46
\[\left \{\left \{y(x)\to -\frac {\sqrt {b} x \tan \left (\sqrt {a} \sqrt {b} x-\sqrt {a} \sqrt {b} c_1\right )}{\sqrt {a}}\right \}\right \}\] ✓ Maple : cpu = 0.037 (sec), leaf count = 25
\[y \relax (x ) = -\frac {\tan \left (\sqrt {a b}\, \left (x +c_{1}\right )\right ) x \sqrt {a b}}{a}\]
Hand solution
\[ ay^{2}+bx^{2}+xy^{\prime }-y=0 \]
This is Riccati first order non-linear. Let \(y=ux\), hence the above becomes\begin {align*} au^{2}x^{2}+bx^{2}+x\left (u^{\prime }x+u\right ) -ux & =0\\ au^{2}x+bx+u^{\prime }x & =0\\ au^{2}+b+u^{\prime } & =0\\ u^{\prime } & =-au^{2}-b \end {align*}
Which is separable, Hence
\[ \frac {du}{au^{2}+b}=-dx \]
Integrating
\begin {align*} \frac {1}{\sqrt {ab}}\arctan \left (\frac {au}{\sqrt {ab}}\right ) & =-x+C\\ \frac {au}{\sqrt {ab}} & =\tan \left (\sqrt {ab}\left (-x+C\right ) \right ) \\ u & =\frac {\sqrt {ab}}{a}\tan \left (\sqrt {ab}\left (-x+C\right ) \right ) \end {align*}
Therefore
\begin {align*} y & =ux\\ & =x\frac {\sqrt {ab}}{a}\tan \left (\sqrt {ab}\left (-x+C\right ) \right ) \end {align*}
Verification