Question

Let y = f(x) be the solution of $\frac{dy}{dx}=\frac{y}{x}+\tan \frac{y}{x},y\left(1\right)=\pi /2$ then

Answer 1

Putting V = y/x, we obtain
$V+x\frac{dV}{dx}=V+\tan V\Rightarrow \frac{dx}{x}=\cot VdV$
Integrating $\log x=\log \sin V+Const$.
$\Rightarrow x=C\sin y/x$.
Putting $x=1,y=\pi /2$, we have C = 1.
Thus $x=\sin \left(\frac{y}{x}\right)\Rightarrow y=x{\sin }^{-1}x=f\left(x\right)$.
f is defined on [-1, 1] and the range of f is $[\frac{-\pi }{2},\frac{\pi }{2}]$.
Clearly f is continuous on its domain which is [-1, 1] and $f\left(x\right)\le \pi /2<2$ for all $xϵ[-1,1]$.