Steps for solution of $\frac{dy}{dx}-y^2=\frac{a}{x}+b$

Question

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

Answer 1

$$\frac{dy}{dx}=y^2+\frac{b}{x}+a$$ This is a Riccati ODE. The usual change of function is : $$y(x)=-\frac{u'(x)}{u(x)}\quad;\quad y'=-\frac{u''}{u}+\frac{(u')^2}{u^2}$$ $$y'=\left(-\frac{u'}{u}\right)^2+\frac{b}{x}+a=-\frac{u''}{u}+\frac{(u')^2}{u^2}$$ $$\frac{b}{x}+a=-\frac{u''}{u}$$ $$u''-\left(\frac{b}{x}+a\right)u=0$$ This is the general confluent hypergeometric differential equation :

https://mathworld.wolfram.com/GeneralConfluentHypergeometricDifferentialEquation.html

From Eq.$(1)$ in this page determine $A$, $f(x)$ and $h(x)$. This leads to :

Then computing $y(x)=-\frac{u'(x)}{u(x)}$ will lead to the solution.

Answer 2

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 $k=-\frac b{2\sqrt{-a}}$.