At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions.

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions

this is what I got so far

after this I am stuck :(

and this is the table I used: http://tutorial.math.lamar.edu/pdf/Laplace_Table.pdf

but what I do about the 4/s^2+4s

factor out the s and that is what I wrote

oh I see; same thing though
factor out s and do partial fractions

but then you get 4/(s(s+4)-->4/s[1/s+4)--> 4e^(-4t)

is that right?

\[\frac4{s^2+4s}=\frac As+\frac B{s+4}\implies A(s+4)+Bs=4\]\[A=1~~~,B=-1\]

oh so you do partial fractions with both of them

that's what I'd do, yeah

and for the other one it's A(s^2+4s) B(s+1) right

\[\mathcal L^{-1}\{\frac1s\}=1\]\[-\mathcal L^{-1}\{\frac1{s+4}\}=-e^{-4t}\]

for the other one it would be...

\[\frac2{s(s+1)(s+4)}=\frac As+\frac B{s+1}+\frac C{s+4}\]

\[A(s+1)(s+4)+Bs(s+4)+Cs(s+1)=2\]

then you get As^2+5As+4A+Bs^2+4Bs+Cs^2+C=2

right

not necessary I don't think
plug in s=-1,-4,and 0 and I think you can get the values quickly

so you dont need ti distribute it?

not necessarily
how did I get the other answers?
I'll show you...

B=-2/3
C=1/6

and A=1/2

thats what I got

I didnt get those values though :(

do you see how I got mine?

i mean for the values of the other one

that I haven't checked yet, one sec...

yay!! so would the final answer be (1/2)-(2/3)(e^-t)+(1/6)e^(-4t)-e^(-4t)

looks good to me :)

it's wrong :(

I think you forgot the one from the other term
do you have infinite tries?

no I ownly have 4 more tries

but what did I forget

only*

ok let me try it again from the top; I went of what you had
I should double-check everything

ok thanks

ok I think I found it...

you dropped the 8...

unfortunately this is going to make partial fractions a real pain...

hyperbolic trig functions*

can you please show me because I honestly have no clue

\[\mathcal L^{-1}\{\frac2{((s+2)^2+2^2)}\}=e^{-2t}\sin(2t)\]

I think you'll have to get A on your own ;)
I'm a bit busy

I think\[(A+C)s^2=0\implies A=-C=-\frac25\]but you shouod double-check me

I think that is correct