Substitutions — Question 3

Question 3 – Solution

Strategy. The substitution is valid for negative $x$ too, but its initial ratio is negative and the logarithm must respect the domain.

Step 1: Transform and integrate. From $y=xv$ and $y'=v+xv'$, $$xv'=\frac 1v,\qquad v\,dv=\frac{dx}{x},\qquad v^2=2\ln|x|+C.$$ At $x=-1$, $v=y/x=-1$, so $C=1$. Continuity and the nonzero ratio select $$v=-\sqrt{1+2\ln|x|}.$$

Step 2: Recover the branch and domain. The radicand must be strictly positive, since zero would give the excluded value $y=0$. Thus $|x|>e^{-1/2}$. On the negative half-line this gives $$\boxed{y=-x\sqrt{1+2\ln|x|},\qquad I=(-\infty,-e^{-1/2})}.$$ It is positive and gives $y(-1)=1$.

Step 3: Verify and check the endpoint. Let $D=1+2\ln|x|$. Since $D'=2/x$, $$y'=-\sqrt D-\frac 1{\sqrt D},\qquad \frac yx+\frac xy=-\sqrt D-\frac 1{\sqrt D}.$$ At the finite endpoint approached from the left, $D\downarrow 0$, so $y\to 0$ from above and $y'\to-\infty$. The original equation is undefined there, and no differentiable extension through it is possible.

Step 4: Identify the two domain mistakes. The real logarithm $\ln x$ is not defined for these negative $x$; the antiderivative is $\ln|x|$. Also $v>0$ would give $y=xv<0$ on this half-line, contradicting $y(-1)=1$. The sign must be selected for the transformed ratio, not guessed from the sign of $y$ alone.