Consider the initial value problem $y''+49y=\cos(7t), \quad y(0)=9, \quad y'(0)=3$. Solve for $y(t)$.
Until now, I have found that $$Y(s)=\frac{s}{(s^2+49)(s^2+49)}+9\frac{s}{s^2+49}+3\frac{1}{s^2+49}$$
I have trouble with the first factor $$\frac{s}{(s^2+49)(s^2+49)}$$ because I can not find a useful way to solve it. By partial fractions I have found nothing useful.
reverse engineering... $\mathcal L \{tf(t)\} = -\frac {d}{ds} \mathcal L\{f(t)\}$
$\frac {s}{(s^2 + 49)^2} = -\frac{d}{ds}\frac 12\frac {1}{s^2 + 49}$