\[ a y(x)^2+b x^2+x y'(x)-y(x)=0 \] ✓ Mathematica : cpu = 0.102646 (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 = 0.031 (sec), leaf count = 25
\[ \left \{ y \left ( x \right ) =-{\frac {x}{a}\tan \left ( \sqrt {ab} \left ( x+{\it \_C1} \right ) \right ) \sqrt {ab}} \right \} \]
\[ ay^{2}+bx^{2}+xy^{\prime }-y=0 \]
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
restart; ode:=a*y(x)^2+b*x^2+x*diff(y(x),x)-y(x)=0; my_sol:=x*sqrt(a*b)/a*tan(sqrt(a*b)*(-x+_C1)); odetest(y(x)=my_sol,ode); 0