Verify an exponential–sine series by two methods

$y^{\prime}=e^x\sin x+e^x\cos x=y+e^x\cos x$. Hence $y^{\prime\prime}=y^{\prime}+e^x\cos x-e^x\sin x=2(y^{\prime}-y)$. Differentiating gives $y^{\prime\prime\prime}=2(y^{\prime\prime}-y^{\prime})$.

At zero, $y=0$, $y^{\prime}=1$, $y^{\prime\prime}=2$, $y^{\prime\prime\prime}=2$. Thus $e^x\sin x=x+x^2+x^3/3+O(x^5)$. The source derivative list labels the last two entries both as third derivatives; the first of those entries is the second derivative. The values and final coefficients are unchanged.

Use $e^x=1+x+x^2/2+\cdots$ and $\sin x=x-x^3/6+\cdots$. Through degree three,

$\begin{aligned}&e^x\sin x\\ & =x+x^2+\left(\frac12-\frac16\right)x^3+\cdots\\ & =x+x^2+\frac{x^3}{3}+\cdots.\end{aligned}$

The omitted fourth-degree coefficient cancels in the complete product; the displayed series is required only through the cubic term.