How to find the particular solution of the differential equation? $$\frac{dy}{dx} = 1 + \frac{y}{x} + \left(\frac{y}{x}\right)^2 , ~~ y(1)=1$$
Thank you
 A: Substituting $$y(x)=xv(x)$$ then we get
$$x\frac{dv(x)}{dx}+v(x)=v(x)^2+v(x)+1$$ and then we obtain
$$\frac{\frac{dv(x)}{dx}}{v(x)^2+1}=\frac{1}{x}$$
Can you finish?
A: We start by setting a variable, $z$ equal to $\frac{y}{x}$
$$\frac{{\rm d}y}{{\rm d}x} = 1 + z + z^2$$
Next, find $\frac{{\rm d}z}{{\rm d}x}$.
$$\frac{{\rm d}z}{{\rm d}x} = \frac{x \frac{{\rm d}y}{{\rm d}x} - y}{x^2} = \frac{1}{x} \frac{{\rm d}y}{{\rm d}x} - \frac{y}{x^2} = \frac{1}{x}\left(\frac{{\rm d}y}{{\rm d}x} - z\right)$$
Now, solve for $\frac{{\rm d}y}{{\rm d}x}$.
$$\frac{{\rm d}z}{{\rm d}x} = \frac{1}{x}\left(\frac{{\rm d}y}{{\rm d}x} - z\right)$$
$$x \frac{{\rm d}z}{{\rm d}x} = \frac{{\rm d}y}{{\rm d}x} - z$$
$$\frac{{\rm d}y}{{\rm d}x} = z + x \frac{{\rm d}z}{{\rm d}x}$$
Substitute this into the original equation.
$$z + x \frac{{\rm d}z}{{\rm d}x} = 1 + z + z^2$$
$$x \frac{{\rm d}z}{{\rm d}x} = 1 + z^2$$
Now, solve this as a separable differential equation.
$$\frac{{\rm d}z}{1 + z^2} = \frac{{\rm d}x}{x}$$
$$\int\frac{{\rm d}z}{1 + z^2} = \int\frac{{\rm d}x}{x}$$
$$\arctan(z) = \ln(x) + C$$
$$z = \tan(\ln(x) + C)$$
Now, we finally substitute $\frac{y}{x}$ for $z$.
$$\frac{y}{x} = \tan(\ln(x) + C)$$
$$y = x\tan(\ln(x) + C)$$
To find $C$, we substitute $x=1$ and $y=1$.
$$1=1\cdot\tan(\ln(1) + C)$$
$$\arctan 1=C$$
$$C=\frac{\pi}{4}$$
Therefore,
$$y = x\tan\left(\ln(x) + \frac{\pi}{4}\right)$$
A: Try $y=tx$ then 
$$t'x+t=1+t+t^2 \implies t'x=1+t^2$$
Which is separable
$$\int \frac {dt}{t^2+1}=\int \frac {dx}x$$
You can also try a solution like $y=ax$ then
$$\frac{dy}{dx} = 1 + \frac{y}{x} + \left(\frac{y}{x}\right)^2$$
$$a=1+a+a^2 \implies a= \pm i \implies y_1=\pm ix$$
And apply the general method for Riccati 's equation 
when a solution is known
$$y=y_1+u$$
And find the general solution of the Riccati's equation
