Intervals of Validity — Question 3

Question 3 – Solution

Strategy. The exponential substitution is invertible only for $u>0$. Its range can restrict the interval before a coefficient singularity is reached.

Step 1: Transform and invert. On the component $x>1$ containing $2$, $$u'=e^y y'=\frac 1{x-1},\qquad u=\ln(x-1)+C.$$ Since $u(2)=2$, $C=2$. Therefore $$\boxed{y=\ln\bigl(2+\ln(x-1)\bigr)}$$ is a real solution only where $2+\ln(x-1)>0$.

Step 2: Find the interval. The positivity condition is $x>1+e^{-2}$. Let $\alpha=1+e^{-2}$; then $$\boxed{I=(\alpha,\infty)}.$$ In particular, $1<\alpha<2$. The solution fails before reaching $x=1$, even though the original right-hand side is smooth at every finite point $(\alpha,y)$.

Step 3: Verify and analyze the ends. Writing $u=2+\ln(x-1)>0$ gives $$y'=\frac 1{(x-1)u}=\frac{e^{-y}}{x-1},\qquad y(2)=\ln 2.$$ As $x\downarrow\alpha$, $u\downarrow 0$ and $y\to-\infty$, preventing a finite continuous extension. As $x\to\infty$, $u\to\infty$ and $y\to\infty$, but this occurs only at an infinite endpoint and does not shorten the interval.

Step 4: Reject an invalid inverse. The transformed function continues on $(1,\infty)$ but becomes negative on $(1,\alpha)$. No real $y$ can have $e^y=u\le 0$.

On that negative portion, $z=\ln|u|$ would satisfy $z'=1/((x-1)u)<0$, whereas $e^{-z}/(x-1)=1/((x-1)|u|)>0$. Thus the absolute-value replacement fails the original equation and cannot repair the lost exponential range.