Steps for solution of $\frac{dy}{dx}-y^2=\frac{a}{x}+b$ According to Wolfram,
$$\frac{dy}{dx}-y^2=\frac{b}{x}+a$$
The equation above has the solution;

Where $a$ and $b$ are constants. $_1F_1(a;b;x)$ and $U(a,b,x)$ are the Kummer confluent hypergeometric function and confluent hypergeometric function of the second kind, respectively.
I don't know how Wolfram got this solution, can anyone here help me please?
Thank you guys

What is making problem for $$\frac{dy}{dx}-y^2=\frac{b}{x}+a$$ is $a$. For $a=0$, you would get the "simple"
$$y=\frac{c_1 J_1\left(2 \sqrt{b} \sqrt{x}\right)+\sqrt{b} \sqrt{x} \left(\left(c_1-2\right)
   J_0\left(2 \sqrt{b} \sqrt{x}\right)-c_1 J_2\left(2 \sqrt{b}
   \sqrt{x}\right)\right)}{2 \left(1-c_1\right) x J_1\left(2 \sqrt{b} \sqrt{x}\right) }$$ which can also write
$$y=-\frac 1x\frac{\, _0\tilde{F}_1(;1;-b x)}{ \, _0\tilde{F}_1(;2;-b x)}$$ where appears the regularized confluent hypergeometric function.
Now, for $a \neq 0$, the "monster"
$$\frac{\sqrt{-a} \left(-c_1 (U(k,0,2 t)-2 U(k,1,2 t))-2 (t+1) \,
   _1F_1(k+1;2;2 t)-2 (k-1) t \, _1F_1(k+1;3;2 t)\right)}{c_1 U(k,0,2
   t)+2 t \, _1F_1(k+1;2;2 t)} $$ where $t=x\sqrt{-a}$ and $k=-\frac b{2\sqrt{-a}}$.
