Question

Consider the differential equation 2y′′+2y′+2y=0 with y(0)=−1 , y′(0)=−3. Taking the Laplace transform and solving for L(y) yields :
L(y)=F(s) where F(s)=

Answer 1

I get F(s) = \[\frac{ -2s-8 }{ 2s ^{2}+2s + 2 }\] for an answer but it is wrong

Answer 2

Hmmmm it looks correct...
Everything has a 2 in it though, do they want it simplified maybe?

Answer 3

split the fraction up and simplify ... before inverse laplace?

Answer 4

\[2y''+2y'+2y=0,\qquad y(0)=-1,\quad y'(0)=3\\~\\
\quad\color{gray}{2\Big(}y''+y'+y\color{gray}{\Big)}=0\\
~\\
\left[s^2Y-sy(0)-y'(0)\right]+\Big[sY-y(0)\Big]+\Big[Y\Big]=0\\
Y(s^2+s+1)-y(0)(s+1)-y'(0)=0\\
Y(s^2+s+1)+(s+1)+3=0\\
Y(s^2+s+1)=-(s+4)\\
Y=-\frac{s+4}{s^2+s+1}=-\frac{\color{gray}{2\big(}s+4\color{gray}{\big)}}{\color{gray}{2\big(}s^2+s+1\color{gray}{\big)}}\\
\]

Answer 5

oh I see where this is going... cancel the 2's out

Answer 6

and then do partial fractions b^2-4ac will tell you whether or not s^2+s+1 can be factorable

Answer 7

ok so it's a complex root...it's best to leave the denominator alone or you'll be in a mess