Substitutions — Question 8

Question 8 – Solution

Strategy. A reciprocal change of the independent variable absorbs $x^2$, while converting a finite endpoint into an infinite one.

Step 1: Transform the derivative. Since $t=1/x$, $dt/dx=-1/x^2$. Writing $y(x)=Y(t)$ gives $$x^2y'=-Y'(t),\qquad -Y'+Y=e^{-t}.$$ Thus $\boxed{Y'-Y=-e^{-t}}$ for $t>0$. The minus sign comes from the reversal of the independent variable.

Step 2: Solve and return to $x$. The integrating factor is $e^{-t}$: $$(e^{-t}Y)'=-e^{-2t},\qquad Y=\frac 12e^{-t}+Ce^t.$$ Therefore all solutions on the stated half-line are $$\boxed{y(x)=\frac 12e^{-1/x}+Ce^{1/x}}.$$

Step 3: Apply the endpoint condition. As $x\downarrow 0$, $t\to+\infty$. The first term tends to zero, whereas any nonzero $C$ makes the second term unbounded. Consequently the unique bounded choice is $$\boxed{y_b(x)=\frac 12e^{-1/x}}.$$ Its limits are $0$ as $x\downarrow 0$ and $1/2$ as $x\to\infty$, and $\boxed{y_b(1)=1/(2e)}$.

Step 4: Verify and interpret the reversal. Differentiation gives $y_b'=e^{-1/x}/(2x^2)$, so $x^2y_b'+y_b=e^{-1/x}$. The condition near $x=0$ becomes a condition at $t=\infty$, not at $t=0$. Conversely, $x\to\infty$ corresponds to $t\downarrow 0$. The limit at $x=0$ does not make that point part of the stated domain of the equation.