 Solution

Use the slope substitution for equations involving x y'-y: put $v=y/x$ on the positive interval containing the initial point. Then $x{y}^{\prime }-y=x^2{v}^{\prime }$, so $x^2{v}^{\prime }=\cot v$. Near $v(1)=\pi /4$, separating variables gives $\tan vdv=dx/x^2$. Integration yields $\ln |\cos v|=1/x+C$. The initial value fixes the sign and constant, givingChoose the inverse cosine branch that agrees with $v(1)=\pi /4$. The solution isOn this interval the argument lies strictly between zero and one and the original cotangent is defined. At the finite lower endpoint, $v\to 0$ and the cotangent becomes singular, so that endpoint is excluded from the maximal interval of the initial value problem. The slope substitution here is valid without classifying the original equation as homogeneous in the usual $y^{\prime }=F(y/x)$ sense.