Linear Equations — Question 9

Question 9 – Solution

Strategy. Normalize only where $x\ne 0$. To cross the exceptional point, return to the original equation and check value and derivative matching.

Step 1: Solve on the positive half-line. Dividing by $x>0$ gives $y'-y/x=x$. The integrating factor is $e^{-\ln x}=1/x$, so $$(y/x)'=1,\qquad y=x^2+Cx.$$ Since $y(1)=2$, $C=1$, giving $\boxed{y=x^2+x\quad(x>0)}$.

Step 2: Unique differentiable extension. The polynomial $x^2+x$ satisfies $xy'-y=x^2$ everywhere, including $x=0$. Any solution on $x<0$ also satisfies $(y/x)'=1$, hence has form $x^2+Dx$. Continuity at $0$ forces $y(0)=0$, and differentiability requires the slopes at zero to match: $D=1$. Thus $$\boxed{y=x^2+x\text{ is the unique differentiable extension to }\mathbb R.}$$

Step 3: Data only at the exceptional point. With only $y(0)=0$, the same argument allows any common slope $C$ on the two sides. All global differentiable solutions are $$\boxed{y=x^2+Cx,\qquad C\in\mathbb R.}$$ Indeed $x(2x+C)-(x^2+Cx)=x^2$, and every such polynomial has value zero at $0$. This IVP is therefore nonunique.

Step 4: Incompatible data and normalization. At $x=0$, the original equation requires $-y(0)=0$, ruling out $y(0)=1$. The normalized coefficient $-1/x$ is undefined at zero, so regular standard-form conclusions cannot be applied there. The algebraic condition at the exceptional point must be checked separately.