My problem is:
Solve $x'= $$\dfrac{1}{\sin\left(\dfrac{x(t)}{t}\right)}+\dfrac{x(t)}{t}\;,\;$
where $\;x(2) =\dfrac{2\pi}{3}\;.$
First, I did a substitution: $\;y(t)=\dfrac{x(t)}{t}$ $$ x(t)=ty(t)\implies x'(t) = y(t) + ty'(t) $$ $$ $$ $$ y(t) + ty'(t) = y(t) + \frac{1}{\sin(y(t))}$$ $$ y(2) = \frac{x(2)}{2} = \frac{\frac{2\pi}{3}}{2} = \frac{\pi}{3} $$ $$ $$ $$ y'(t) = \frac{1}{t\sin(y(t))} $$ $$ y(2)= \frac{\pi}{3} $$ $$ $$ $$G(t)=\int \frac{1}{t} dt = \ln(t) + C$$ $$H(y)= \int\sin(y) dy = -\cos(y) + C $$ $$H^{-1}(a)= -\arccos(a) $$ $$ x(t) = ty(t) = -t\arccos(\ln(t)) $$
I think I am doing something wrong, and some things are missing, like the invervals. How can I determine the intervalls? And what am I doing wrong?
I appreciate any kind of help.
The constants (in your $H$ and $G$) are not necessarily equal when you integrate. You can group them together into a single constant: $$\frac{dy}{dt}=\frac{1}{t\sin(y)}\quad\Rightarrow\quad\int\sin(y)dy=\int\frac{dt}{t}\quad\Rightarrow\quad -\cos(y)=\ln(t)+C$$
Applying the initial condition gives $C=-\ln(2)-1/2$, so $$\cos(y)=\ln\left(\frac{2}{t}\right)+\frac{1}{2}\quad\Rightarrow\quad y(t)=\arccos\left(\ln\left(\frac{2}{t}\right)+\frac{1}{2}\right).$$