Show that the function y(x) = (x)/(e^x-1), x ≠ 0 4pt] 1, x=0…

Show that the function \[y(x) = \begin{cases}\dfrac{x}{e^x-1}, & x \neq 0\\[4pt] 1, & x=0\end{cases}\] satisfies a first-order differential equation of the form $y'=R(x,y)$, where $R$ is a rational function in $x$ and $y$.

Explanation of Method

For $x\neq 0$, differentiate with the quotient rule, then eliminate the transcendental term $e^x$ using the relation $e^x=(x+y)/y$ obtained by solving $y=x/(e^x-1)$ for $e^x$. The removable point $x=0$ is then checked separately using a Taylor/L'Hôpital limit argument.

Differentiate for $x \neq 0$

\[y' = \frac{(e^x-1) - xe^x}{(e^x-1)^2} = \frac{e^x(1-x)-1}{(e^x-1)^2}.\]

Express $e^x$ in terms of $x,y$

From $y(e^x-1)=x$: \[e^x = \frac{x+y}{y}, \qquad e^x-1 = \frac{x}{y}.\]

Substitute and simplify

\[y' = \frac{\frac{x+y}{y}(1-x)-1}{(x/y)^2} = \frac{[(x+y)(1-x)-y]\,y}{x^2}.\] Expanding $(x+y)(1-x)-y = x-x^2-xy = x(1-x-y)$, so \[y' = \frac{x(1-x-y)y}{x^2} = \frac{y(1-x-y)}{x}.\]

State the rational ODE

\[\boxed{y' = \frac{y(1-x-y)}{x}} \quad\Longleftrightarrow\quad xy' = y(1-x-y),\] which is manifestly rational in $x,y$. Back-substituting confirms this reproduces $y'=\dfrac{e^x(1-x)-1}{(e^x-1)^2}$ exactly for $x\neq0$.

Continuity of $y$ at $x=0$

By L'Hôpital (or the Taylor expansion of $e^x$), $\lim_{x\to 0}\frac{x}{e^x-1} = \lim_{x\to0}\frac{1}{e^x}=1$, matching the defined value $y(0)=1$, so $y$ is continuous at $x=0$.

Resolving the $0/0$ limit at $x=0$ (completed)

The source left the value of $\lim_{x\to0} y'$ unverified, merely asserting that "L'Hôpital's rule confirms" the equation without computing it. This was completed here using the Maclaurin (Bernoulli-number) series \[y(x) = \frac{x}{e^x-1} = 1 - \frac{x}{2} + \frac{x^2}{12} - \frac{x^4}{720} + \cdots\] Two independent computations were carried out and cross-checked:

(1) Direct differentiation: differentiating the series (or using $\text{sympy.diff}$ and $\text{sympy.limit}$ on the exact function) gives $y'(x) = -\tfrac12 + \tfrac{x}{6} - \tfrac{x^3}{180}+\cdots$, so \[\lim_{x\to0} y'(x) = -\frac12.\]

(2) Substitution into the ODE's right-hand side: plugging the series for $y$ into $R(x,y)=\dfrac{y(1-x-y)}{x}$ (a $0/0$ form at $x=0$, since $y(0)(1-0-y(0))=1\cdot0=0$) and expanding gives \[\frac{y(1-x-y)}{x} = -\frac12 + \frac{x}{6} - \frac{x^3}{180}+\cdots \ \xrightarrow[x\to0]{}\ -\frac12.\]

Both methods agree exactly: $y'(0) = -\tfrac12$, so the differential equation $y'=\dfrac{y(1-x-y)}{x}$ is satisfied (in the limiting/removable sense) at $x=0$ as well, with the specific numerical value $y'(0)=-\tfrac12$ — closing the gap the source left open.

Result

Final answer:

Differential equation (valid for all $x$, including $x=0$ by continuity): \[\boxed{y' = \frac{y(1-x-y)}{x}}\] with $R(x,y)=\dfrac{y(1-x-y)}{x}$ rational in $x,y$. At the removable point, $y(0)=1$ and $\displaystyle\lim_{x\to0}y'(x) = -\frac12$ (verified two independent ways: differentiating the series for $y$ directly, and substituting that series into the ODE's right-hand side), so the equation genuinely holds at $x=0$ too — a claim the source asserted but never actually computed.