Answer:
Step-by-step explanation:
DE given is
y''+8y'+15y=0, y(0)=0, y'(0)=1
Take Laplace on the DE
We get
[tex]s^2 Y(s) -sY(0) -y'(0)+8(sY(s)-y(0))+15Y(s) =0\\s^2 Y(s) -s(0) -1+8(sY(s)-0)+15Y(s) =0\\Y(s)(s^2+8s+15)-1=0\\Y(s) = \frac{1}{s^2+8s+15}[/tex]
Simplify to get
Y(s) = [tex]\frac{1}{2}[ {\frac{1}{s+3} -\frac{1}{s+5} }][/tex]
Take inverse
[tex]y(t) = \frac{e^{-3t} -e^{-5t}}{2}[/tex]
