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

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,
$$y = x\tan\left(\ln(x) + \frac{\pi}{4}\right)$$
