Re: Laplace Transforms
1) y'' - 6y' +13y = 0 ; y(0)=0, y' (0)= -3.
I will step through this one so you can do the others on your own. OK?
LaPlace transforms turn a DE into an algebraic, so we can solve easier.
\(\displaystyle y'=pY-y(0)\)
\(\displaystyle y''=p^{2}Y-py(0)-y'(0)\)
Sub these into your DE:
\(\displaystyle p^{2}Y-py(0)-y'(0)-6(pY-y(0))+13Y=0\)
Sub in the initial conditions given, factor out Y, and get:
\(\displaystyle Y(p^{2}-6p+13)=-3\)
\(\displaystyle Y=\frac{-3}{p^{2}-6p+13}\)
Now, look this up in a table to LaPlace transforms and we get:
\(\displaystyle \boxed{y=\frac{-3}{2}e^{3t}sin(2t)}\)
For the last one, find the laplace of t^2. Which is \(\displaystyle \frac{2}{p^{3}}\)
\(\displaystyle pY-y(0)+5Y=\frac{2}{p^{3}}\)
\(\displaystyle pY+5Y=\frac{2}{p^{3}}\)
\(\displaystyle Y=\frac{2}{p^{3}(p+5)}\)
Taking partial fractions:
\(\displaystyle Y=\frac{-2}{125(p+5)}+\frac{2}{125p}-\frac{2}{25p^{2}}+\frac{2}{5p^{3}}\)
Now, these are relatively easier ones by looking them up in the Laplace table.