Verify an implicit logarithm expansion by a second method

This is labelled Self-Reading, and both copies already print the complete solution. From $e^y=2+\sin2x$, differentiating gives $e^yy^{\prime}=2\cos2x$. Differentiate that product once more:

$e^yy^{\prime\prime}+e^y(y^{\prime})^2=-4\sin2x,$

which is the required identity.

Both copies retain the uncollected third-derivative line:

$\begin{aligned}&e^yy^{\prime\prime\prime}+e^yy^{\prime\prime}y^{\prime}+2e^yy^{\prime\prime}y^{\prime}\\ &{}+e^y(y^{\prime})^3\\ & =-8\cos2x.\end{aligned}$

Collecting the two mixed terms gives

$\begin{aligned}&e^yy^{\prime\prime\prime}+3e^yy^{\prime\prime}y^{\prime}+e^y(y^{\prime})^3\\ & =-8\cos2x.\end{aligned}$

The source values are $y(0)=\ln2$, $y^{\prime}(0)=1$, $y^{\prime\prime}(0)=-1$, $y^{\prime\prime\prime}(0)=-2$. Independently, they follow successively from $e^{y(0)}=2$, $2y^{\prime}(0)=2$, $2y^{\prime\prime}(0)+2=0$, and $2y^{\prime\prime\prime}(0)-6+2=-8$.

The complete printed substitution and simplified series are

$\begin{aligned}&y\\ & =\ln2+x+\frac{-1}{2!}x^2+\frac{-2}{3!}x^3+\cdots\\ & =\ln2+x-\frac12x^2-\frac13x^3+\cdots.\end{aligned}$

Both sources show the independent standard-series route:

$\begin{aligned}&y\approx\ln\left[2+2x-\frac{(2x)^3}{6}\right]\\ & =\ln\left[2\left(1+x-\frac23x^3\right)\right]\\ & =\ln2+\ln\left(1+x-\frac23x^3\right).\end{aligned}$

Substituting the new logarithm argument gives

$\begin{aligned}&y\\ & =\ln2+\left(x-\frac23x^3\right)\\ &{}-\frac{(x-\frac23x^3)^2}{2}\\ &{}+\frac{(x-\frac23x^3)^3}{3}+\cdots.\end{aligned}$

Retaining through the cubic term,

$y\approx\ln2+x-\frac23x^3-\frac12x^2+\frac13x^3=\ln2+x-\frac12x^2-\frac13x^3+\cdots,$

verifying part (ii). These substitutions compare retained coefficients near zero; replacing sine by its finite truncation is not an exact functional identity.

Read the square root in the denominator carefully: the reciprocal of $\sqrt{1-x}$ is $(1-x)^{-1/2}$. Both copies give the full chain

$\frac{\ln(2+\sin2x)}{\sqrt{1-x}}\approx\left(\ln2+x-\frac12x^2-\frac13x^3\right)(1-x)^{-1/2}.$

Only terms through $x^2$ are now required. The retained binomial terms are

$\begin{aligned}&1+\frac12x\\ &{}+\frac{(-1/2)(-3/2)}{2!}(-x)^2\\ & =1+\frac12x+\frac38x^2.\end{aligned}$

Multiplying the two retained quadratic polynomials gives

$\left(\ln2+x-\frac12x^2\right)\left(1+\frac12x+\frac38x^2\right)$

$\approx\ln2+x-\frac12x^2+\left(\frac12\ln2\right)x+\frac12x^2+\left(\frac38\ln2\right)x^2$

$=\boxed{\ln2+\left(1+\frac12\ln2\right)x+\left(\frac38\ln2\right)x^2}.$

The two non-logarithmic quadratic contributions cancel. The omitted terms begin at cubic order; the binomial factor is used near zero with $|x|<1$ and the real denominator requires x<1.