How to find the particular solution of the differential equation?

Question

$$\frac{dy}{dx} = 1 + \frac{y}{x} + \left(\frac{y}{x}\right)^2 , ~~ y(1)=1$$

Thank you

Answer 1

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

Answer 2

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?

Answer 3

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