Question

Use the Laplace transform and the inverses to solve the given initial value problem
\[y''-2y'+5y=0 ; y(0)=1, y'(0)=3\]

Answer 1

So \( L\{y"\}-2L\{y'\}+5L\{y\}=L\{0\}\)

\(s^2Y(s)-sy(0)-y'(0)-2(sY(s)-y(0))+5Y(s)=0\)

\(Y(s)(s^2-2s+5)=s+1\)

\(Y(s)=\large \frac{s+1}{s^2-2s+5}\)

How do I continue from here?

Answer 2

the bottom becomes:
\[Y(s)=\frac{s+1}{(s^2-2s+1+4)}=\frac{s-(-1)}{(s-1)^2+2^2}\]You might be off by a sign on top. If that was s-1, this would fit into \[e^{at}Cos(bt)\]

Answer 3

http://www.sosmath.com/diffeq/laplace/table/table.html

Answer 4

where a=1, and b=2

Answer 5

THANKKKKSSS :)