Solve by the method of variation of parameters y'' - y =…

Solve by the method of variation of parameters $y'' - y = \frac{2e^x}{e^x-1}$.

Step 1: Solve the homogeneous equation

\[ r^2-1=0 \implies r=\pm1, \qquad y_1=e^x,\ \ y_2=e^{-x}. \]

Step 2: Wronskian

\[ W(y_1,y_2) = \begin{vmatrix} e^x & e^{-x} \\[4pt] e^x & -e^{-x} \end{vmatrix} = -1-1 = -2. \]

Step 3: Set up the variation-of-parameters system

Write $y=u_1e^x+u_2e^{-x}$, with $f(x)=2e^x/(e^x-1)$. By Cramer's rule ($W=-2$): \[u_1' = -\frac{y_2f}{W} = \frac{1}{e^x-1}, \qquad u_2' = \frac{y_1f}{W} = -\frac{e^{2x}}{e^x-1}.\]

Step 4: Integrate

For $u_1$, substitute $t=e^x$, $dx=dt/t$: \[u_1 = \int\frac{dx}{e^x-1} = \int\frac{dt}{t(t-1)} = \int\!\left(\frac{1}{t-1}-\frac1t\right)dt = \ln|t-1|-\ln t = \ln(e^x-1) - x.\]

For $u_2$, the same substitution gives $e^{2x}/(e^x-1)=t^2/(t-1)=t+1+1/(t-1)$, i.e. $e^{2x}/(e^x-1)=e^x+1+1/(e^x-1)$, so

\[ u_2 = -\int\!\left(e^x+1+\frac{1}{e^x-1}\right)dx = -e^x - x - \int\frac{dx}{e^x-1}. \]

The last integral is exactly $u_1$ above — namely $\ln(e^x-1)-x$, not just $\ln(e^x-1)$. Substituting this in: $u_2 = -e^x - x - \bigl[\ln(e^x-1)-x\bigr] = -e^x - \ln(e^x-1).$

Independent check (differentiate directly, no substitution): with $u_1=\ln(e^x-1)-x$,

\[ u_1' = \frac{e^x}{e^x-1} - 1 = \frac{e^x-(e^x-1)}{e^x-1} = \frac{1}{e^x-1}.\ \checkmark \]

and with $u_2=-e^x-\ln(e^x-1)$, \[u_2' = -e^x - \frac{e^x}{e^x-1} = -\frac{e^x(e^x-1)+e^x}{e^x-1} = -\frac{e^{2x}}{e^x-1}.\ \checkmark\]

A direct re-integration of $u_2$ is worth flagging: it is easy to reuse the antiderivative $\ln(e^x-1)$ for $\int dx/(e^x-1)$ and forget the $-x$ term found while computing $u_1$ — that slip produces a spurious extra $-x$ inside $u_2$, which does not cancel correctly in the final check. The computation above keeps the two consistent.

Step 5: Form the particular solution

\[ y^* = u_1e^x + u_2e^{-x} = \bigl[\ln(e^x-1)-x\bigr]e^x + \bigl[-e^x-\ln(e^x-1)\bigr]e^{-x}, \]

\[ = e^x\ln(e^x-1) - xe^x - 1 - e^{-x}\ln(e^x-1) = (e^x-e^{-x})\ln(e^x-1) - xe^x - 1. \]

Step 6: Verify $y^*$

Split $y^*=\varphi+\psi+c$ with $\varphi=(e^x-e^{-x})\ln(e^x-1)$, $\psi=-xe^x$, $c=-1$, and write $t=e^x$. With $A=e^x-e^{-x}$, $B=\ln(e^x-1)$: $A'=e^x+e^{-x}$, $A''=A$, $B'=e^x/(e^x-1)$, $B''=-e^x/(e^x-1)^2$, so

\[ \varphi''-\varphi = 2A'B' + AB'' = \frac{2(t^2+1)}{t-1} - \frac{t+1}{t-1} = \frac{2t^2-t+1}{t-1}. \]

For $\psi=-xe^x$: $\psi'=-e^x(1+x)$, $\psi''=-e^x(2+x)$, so

\[ \psi''-\psi = -e^x(2+x) + xe^x = -2e^x = -2t. \]

and $c''-c = 0-(-1)=1$. Summing all three, over the common denominator $t-1$: \[y^{*\prime\prime}-y^* = \frac{2t^2-t+1}{t-1} - 2t + 1 = \frac{2t^2-t+1-(2t-1)(t-1)}{t-1} = \frac{2t^2-t+1-(2t^2-3t+1)}{t-1} = \frac{2t}{t-1}.\]

Since $t=e^x$, this is exactly $2e^x/(e^x-1). $\checkmark$

Step 7: General solution

The solution is recorded in the result below.