The reason is the little \$t\$ that sits in front of the sine function there. First set up the equation in standard form:
$$\frac{\text{d}}{\text{d}t} v_t+\frac12 v_t=2.5 t\sin t$$
This can also be expressed a little more elegantly as:
$$\left[D +\frac12\right] v_t=2.5 t\sin t$$
The denominator of the Laplace transform of the right side is \$\left(s^2+1\right)^2\$. So \$\left[\left(D^2+1\right)^2\right]\$ will annihilate the right expression. The fact that this is squared means that terms for the first power of \$t\$ must be included in the solution.
Here's proof of that:
$$\begin{align*} \left[\left(D^2+1\right)^2\right]2.5 t\sin t \\ \left[D^2+1\right]\left[D^2+1\right]2.5 t\sin t \\ \left[D^2+1\right]\left[\vphantom{D^2+1}\left(-2.5 t\sin t+5\cos t\right)+2.5 t\sin t\right] \\ \left[D^2+1\right]5\cos t \\ \left[\left(-5\cos t\right)+5\cos t\right] \\ 0 \end{align*}$$
Note that it took two of the \$\left[D^2+1\right]\$ to kill both the \$t\$ and also the \$\sin t\$. That's why it had to be squared. If there was a \$t^2\$ in front, a third power would have been required, instead. Etc.
So the original equation becomes:
$$\begin{align*}\left[D +\frac12\right] v_t&=2.5 t\sin t\\\\ \left[D^2+1\vphantom{\frac12}\right]^2\left[D +\frac12\right] v_t&=\left[D^2+1\vphantom{\frac12}\right]^22.5 t\sin t\\\\\left[D^2+1\vphantom{\frac12}\right]^2\left[D +\frac12\right] v_t&=0\end{align*}$$
That's homogeneous and will have the general solution form of:
$$v_t=A_1\exp\left(-\frac12 t\right)+A_2 t\cos t+A_3 t\sin t+A_4\cos t+A_5\sin t$$
The first term (the exponential) provides the general solution to the associated homogeneous equation and the last four constitute a particular solution to the non-homogeneous part, with undetermined coefficients.
Direct substitution results in:
$$\begin{align*}\left[D+\frac12\right]v_t&=\left(\frac12A_2+A_3\right)t\cos t\\&+\left(-A_2+\frac12 A_3\right)t\sin t\\&+\left(A_2+\frac12 A_4+A_5\right)\cos t\\&+\left(A_3-A_4+\frac12 A_5\right)\sin t\\&=2.5 t\sin t\end{align*}$$
The undetermined coefficients can now be worked out simultaneously to yield:
$$v_t=A_1\exp\left(-\frac12 t\right)-2 t\cos t+ t\sin t+\frac85\cos t+\frac65\sin t$$
Plug that back into the standard form shown at the outset above and find that the equation now balances.
See Section 6.3, "Undetermined Coefficients and the Annihilator Method" from the 9th edition of "Fundamentals of Differential Equations" by Nagle, Saff, and Snider, Pearson 2018
