Question

Y'=y/x-ctg(y/x) решите дифференциальные уравнения solve the following differential equations

Answer 1

$y'=\frac{y}{x}-ctg(\frac{y}{x})$
$\frac{y}{x}=t$
$y=tx$
$y'=t'x+t$

$t'x+t=t-ctg t$
$\frac{dt}{dx} x=-ctg t$
$\frac{dt}{ctg t} =- \frac{dx}{x}$
$\frac{sint dt}{cost}= - \frac{dx}{x}$
$-\frac{d(cost)}{cost}= - \frac{dx}{x}$
$ln cos t=lnx+C$
$cost=Cx$
$t=arccos(Cx)$
$\frac{y}{x}=arccos(Cx)$
$y=x*arccos(Cx)$