Homogeneous differential equations

The equation is already of the form $F\left(\frac{y}{x}\right)$ with $F(v)=v+1$. With $y=vx$: $$v+x\frac{\mathrm{d}v}{\mathrm{d}x}=v+1\Rightarrow x\frac{\mathrm{d}v}{\mathrm{d}x}=1\Rightarrow \mathrm{d}v=\frac{\mathrm{d}x}{x}.$$ Integrating: $v=\ln x+C$. Undo the substitution: $y=x(\ln x+C)$. With $y=2$ at $x=1$: $2=1\cdot (0+C)$, so $C=2$. $$y=x(\ln x+2).$$

Divide the numerator and denominator by $x^2$: $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1+3v^2}{2v}$ with $v=\frac{y}{x}$. With $y=vx$: $$v+x\frac{\mathrm{d}v}{\mathrm{d}x}=\frac{1+3v^2}{2v}\Rightarrow x\frac{\mathrm{d}v}{\mathrm{d}x}=\frac{1+3v^2-2v^2}{2v}=\frac{1+v^2}{2v}.$$ Separate the variables: $\frac{2v}{1+v^2}\mathrm{d}v=\frac{\mathrm{d}x}{x}$. Integrating: $\ln (1+v^2)=\ln x+C$, that is, $1+v^2=Kx$ with $K={\mathrm{e}}^C$.

Undo the substitution: $1+\frac{y^2}{x^2}=Kx$. With $y=1$ at $x=1$: $2=K$. So $y^2=2x^3-x^2$ and, as $y>0$, $$y=x\sqrt{2x-1}.$$

a) Dividing by $x^2$: $\frac{\mathrm{d}y}{\mathrm{d}x}=1+v+v^2$, with $v=\frac{y}{x}$. As $\frac{\mathrm{d}y}{\mathrm{d}x}=v+x\frac{\mathrm{d}v}{\mathrm{d}x}$: $$v+x\frac{\mathrm{d}v}{\mathrm{d}x}=1+v+v^2\Rightarrow x\frac{\mathrm{d}v}{\mathrm{d}x}=1+v^2.$$

b) Separate: $\frac{\mathrm{d}v}{1+v^2}=\frac{\mathrm{d}x}{x}$, and integrate: $\arctan v=\ln x+C$. With $x=1$, $v=\sqrt{3}$: $\arctan \sqrt{3}=\frac{\pi }{3}=\ln 1+C$, so $C=\frac{\pi }{3}$. Then $v=\tan \left(\ln x+\frac{\pi }{3}\right)$ and $$y=x\tan \left(\ln x+\frac{\pi }{3}\right).$$

c) With $x={\mathrm{e}}^{-\pi /12}$: $\ln x=-\frac{\pi }{12}$ and $-\frac{\pi }{12}+\frac{\pi }{3}=\frac{\pi }{4}$. So $y={\mathrm{e}}^{-\pi /12}\tan \frac{\pi }{4}={\mathrm{e}}^{-\pi /12}$.

In the exam: in a) marks are given for writing the derivative of $y=vx$ as $v+x\frac{\mathrm{d}v}{\mathrm{d}x}$ and reaching the required result without skipped steps. In b), undoing the substitution ($y=xv$) and using the initial condition are two separate marking points.