y"+3y'+2y=u2(t) and y(0)=0 and y'(0)=1
y"-y=20δ(t-3) and y(0)=0 and y'(0)=0
I am having a bit of trouble with these problems. Can anyone help?
Ben
Alright here is the first one
$\displaystyle s^2Y-s(0)-1+3sY+2Y=\frac{1}{s}$
First we need to isolate Y
$\displaystyle s^2Y-s(0)-1+3sY+2Y=\frac{1}{s} \iff (s^2+3s+2)Y=s + 1 +\frac{1}{s} $
$\displaystyle Y =\frac{s+1}{(s^2+3s+2)}+\frac{1}{s(s^2+3s+2)}$
Factoring and reducing like terms we get
$\displaystyle Y=\frac{s+1}{(s+1)(s+2)}+\frac{1}{s(s+1)(s+2)}=\fr ac{1}{s+2}+\frac{1}{s(s+1)(s+2)}$
Depending on what theorems you know there are many ways to take the inverse Laplace transform, but lets do partial fractions on the 2nd fraction
(Work not shown)
$\displaystyle \frac{1}{s(s+1)(s+2)}=\frac{1/2}{s}+\frac{-1}{(s+1)}+\frac{1/2}{s+2}$
Putting this into the above and collecting like terms we get
$\displaystyle Y=\frac{1}{2}\left( \frac{1}{s} \right)-\frac{1}{s+1}+\frac{3}{2}\left( \frac{1}{s+2}\right)$
Now taking the inverse transform we get
$\displaystyle y(t)=\frac{1}{2}-e^{-t}+\frac{3}{2}e^{-2t}$
Try the 2nd one and show us what you have done!!
Good luck.
If you take two derivatives and plug in back in you can check your work. However, this solution doesn't check. It doesn't satify the inital condition aswell.
$\displaystyle s^2Y-Y=20e^{-3s} \iff (s^2-1)Y=20e^{-3s} \iff Y =20e^{-3s}\frac{1}{(s-1)(s+1)}$
By partial fractions or this trick
$\displaystyle \frac{1}{(s-1)(s+1)}=\frac{\frac{1}{2}s-\frac{1}{2}s+\frac{1}{2}+\frac{1}{2}}{(s-1)(s+1)}=\frac{\frac{1}{2}s+\frac{1}{2}}{(s+1)(s-1)}+\frac{-\frac{1}{2}s+\frac{1}{2}}{(s+1)(s-1)}$
$\displaystyle \frac{\frac{1}{2}(s+1)}{(s+1)(s-1)}+\frac{-\frac{1}{2}(s-1)}{(s+1)(s-1)}=\frac{\frac{1}{2}}{(s+1)}-\frac{\frac{1}{2}}{(s-1)}$
Now we get
$\displaystyle Y=\frac{10e^{-3s}}{(s+1)}-\frac{10e^{-3s}}{(s-1)}$
Now taking the inverse Laplace transform we get
$\displaystyle y(t)=10e^{-(t-3)}\mathcal{U}(t-3)-10e^{(t-3)}\mathcal{U}(t-3)$