I am going to use y instead of Q.
Are you familiar with Laplace transforms?. They are a cool way to solve a second order DE given initial conditions.
\(\displaystyle y''+300y'=-1000, \;\ y(0)=2, \;\ y'(0)=2\)
Take the Laplace of -1000 and rewrite using the transforms:
\(\displaystyle L(-1000)=\frac{-1000}{p}\)
\(\displaystyle p^{2}Y-2p+300(pY-2)=\frac{-1000}{p}\)
\(\displaystyle Y=\frac{-1}{90(p+300)}+\frac{181}{90p}-\frac{10}{3p^{2}}\)
Now, if we use the inverse Laplace on this, we get our solution complete with initial conditions implemented.
\(\displaystyle y=\frac{-1}{90}e^{-300t}-\frac{10}{3}t+\frac{181}{90}\)
These are usually looked up in a table or done with tech of some sort.
I have to go, so I do not have tinme now to explain further. But let me know if you need more assistance.
