Find a solution for the initial-value problem $\frac{dx}{dt}t(\frac{dx}{dt}+2) = x$ with homogeneous initial conditions, and give a possible interval of definition of this solution.
Progress I have made so far: Using the substitutions such as $u=x+t$ or $u=\frac{x+t}{t}$ but no productive results. Any help will be kindly appreciated.
On
$$\frac{dx}{dt}(\frac{dx}{dt}+2) = \frac{x}{t}$$ Completing the square in the above expression, $$\frac{dx}{dt} = \pm\sqrt{\frac{x}{t}+1}-1$$ Considering the positive root first,
Let $x=ut$ then, $\frac{dx}{dt} = u + t\frac{du}{dt}$ $$\therefore u+t\frac{du}{dt}=\sqrt{u+1}-1$$ $$\therefore \frac{du}{\sqrt{1+u}-(1+u)} = \frac{dt}{t}$$ $$\therefore \frac{du}{\sqrt{1+u}(1-\sqrt{1+u)}} = \frac{dt}{t}$$
This can be integrated to obtain $u$ in terms of $t$. Finally back substitute to get $x(t)$
Similarly, we can solve for the other root. $$\therefore u+t\frac{du}{dt}=-\sqrt{u+1}-1$$ $$\therefore \frac{du}{\sqrt{1+u}+(1+u)} = -\frac{dt}{t}$$ $$\therefore \frac{du}{\sqrt{1+u}(1+\sqrt{1+u)}} = -\frac{dt}{t}$$
Hope this helps.
$$\frac{dx}{dt}t \left(\frac{dx}{dt}+2 \right) = x$$ $$x=x't(x'+2)$$ Rewrite it as: $$x=t(x'^2+2x')$$ This is d'Alembert's equation. $$x=tf(x') \text { where } f(x')=x'^2+2x'$$
$$x=t(x'^2+2x')$$ Differentiate: $$-x'=2x'x''t+x'^2+2tx''$$ $$-(x'+2tx'')=x'(2x''t+x')$$ $$(x'+1)(2x''t+x')=0$$ First equation gives us: $$(x'+1) =0\implies x(t)=-t$$ The second equation gives us : $$(2x''t+x')=0 \implies (\ln x')'=-\dfrac 1 {2t}$$ $$\ln x'= -\frac 12 \ln t +c \implies x'=kt^{-1/2}$$ $$x(t)=C_1\sqrt t + C_2$$ Plug this in the original equation to find the constants. $$\implies C_2=\dfrac {C_1^2}4$$ Finally: $$\boxed {x(t)=C_1 \left(\sqrt t + \dfrac {C_1}4 \right )}$$