Question

Let $\frac{xdy}{dx}=y+(x^2-y^2{)}^{1/2},y(1)=0$

Answer 1

$\frac{dy}{dx}=\frac{y}{x}+{[1-{\left(\frac{y}{x}\right)}^2]}^{1/2}$
Putting y = vx, we obtain
$v+\frac{xdv}{dx}=v+(1-v^2{)}^{1/2}$
$\int \frac{1}{(1-v^2{)}^{1/2}}dv=\int \frac{1}{x}dx{\sin }^{-1}v=\log x+C$
Putting $y\left(1\right)=0$, we have $v\left(1\right)=y\left(1\right)/1=0$, and $C={\sin }^{-1}0-\log 1=0$. Hence
$v=\frac{y}{x}=\sin \left(\log x\right)\Rightarrow y=x\sin \left(\log x\right)$
So $y\le x$. Also $f\left(e\right)=e\sin 1\ge e/2$
Morever, $y\left(e^{\pi /2}\right)=e^{x/2}$ so $y\left(e^{\pi /2}\right)/e^{/pi/2}=1ϵ[-1,1]$ and (0, 2). The range of y/x is clearly [-1, 1].