You want find the Laplace Transform of \(\displaystyle \frac{dv}{dt}+ 3v- 13sin(2t)\)
Yes, the Laplace Transform is "linear" so you can take the Laplace transform of each part:
Letting the Laplace Transform of v be "V" the Laplace transform of \(\displaystyle \frac{dv}{dt}\) is \(\displaystyle sV- v(0)= sV- 6\).
The Laplace Transform of \(\displaystyle -13 sin(2t)\) is (either using a table of Laplace transforms or by actually doing the integration by parts twice) \(\displaystyle -13\frac{2}{s^2+ 4}= \frac{-26}{s^2+ 4}\).
So the Laplace Transform of \(\displaystyle \frac{dv}{dt}+ 3v- 13sin(2t)\) is
\(\displaystyle sV- 6+ 3V- \frac{26}{s^2+ 4}\).
In your solution, you appear to have \(\displaystyle L(3v)= 3L(v)= \frac{3}{s^2}\). Since, in your notation, you are using "V" to represent the Laplace transform of v, the Laplace transform of 3v is just 3V. I suspect you used the Laplace transform of independent variable, "t", not the dependent variable, "v".
