Question

Solving an IVP with Laplace-Heaviside, don't need help yet, just writing it out.

Answer 1

Using the Laplace Transform, solve the given Initial Value Problem.

\[y'+2y=f(t), \ \ \ y(0)=0\]where\[f(t) =\begin{cases}t, \ \ \ 0 \leq t<1\\0, \ \ \ t\ge1.\end{cases}\]

Answer 2

First, putting the function f(t) in terms of a linear combination of Heaviside functions:\[f(t)=t-t \ \mathcal{U}(t-1).\]
Plugging in and taking the Laplace transform of both sides,\[\mathcal{L}\left\{y'\right\}+\mathcal{L}\left\{2y\right\}=\mathcal{L}\left\{t-t \ \mathcal{U}(t-1)\right\} = sY(s)-y(0)+Y(s)=\frac{1}{s^2}-\frac{e^{-s}}{s^2}\]

Answer 3

\[Y(s)[s+2]=\frac{1}{s^2}-\frac{e^{-s}}{s^2},\]\[Y(s)=\frac{1-e^{-s}}{s^2(s-1)}\]

Answer 4

Splitting these up into two separate fractions and then taking partial fractions of the first (and applying it to the second one)-, I get\[\frac{1}{s^2(s-1)}=\frac{A}{s}+\frac{B}{s^2}+\frac{C}{(s+1)};\]\[As(s+1)+B(s+1)+Cs^2=1.\]

Answer 5

uh is'nt it\[Y(s)=\frac{ 1-e^{-s}}{ s^2(s+2) }\]??

Answer 6

Yup, yup, my bad, totally missed the typo! Thank you.

Answer 7

yeah and the rest is ok

Answer 8

\[s^2(A+C)+s(2A+B)+2B=1;\]\[B=\frac{1}{2},A=\frac{1}{4}, C=-\frac{1}{4}\]

Answer 9

i'm getting something else

Answer 10

:o

Answer 11

Lul one sec, lemme double check. \[s^2(A+C)=s^2(0); \ A = -C.\]\[s(2A+B)=s(0); \ 2A+B=0.\]OH, right there.

Answer 12

\[B=-\frac{1}{2}A\]

Answer 13

yup

Answer 14

\[2B=1; \ B=\frac{1}{2}\]

Answer 15

Alright, so A = -1/4th, and C = 1/4.

Answer 16

yes !!

Answer 17

That in mind, \[\Bigg[-\bigg(\frac{1}{4}\bigg)\frac{1}{s}+\bigg(\frac{1}{2}\bigg)\frac{1}{s^2}+\bigg(\frac{1}{4}\bigg)\frac{1}{s+2}\Bigg]\]

Answer 18

Alright, from there,\[-\frac{1}{4}\mathcal{L^{-1}}\left\{\vphantom{}\frac{1}{s}\right\}+\frac{1}{2}\mathcal{L^{-1}}\left\{\vphantom{}\frac{1}{s^2}\right\}+\frac{1}{4}\mathcal{L^{-1}}\left\{\vphantom{}\frac{1}{s+2}\right\}\]

Answer 19

yes!!

Answer 20

(I feel like a huge troll taking my time to write all of the proper math, lol, sorry)

Answer 21

nah Latex is the right way to write it out

Answer 22

\[\frac{ -1 }{ 4 }u(t)+\frac{ 1 }{ 2 }t.u(t)+\frac{ 1 }{ 4 }e^{-2t}u(t)\]
seems fine?

Answer 23

\[-\frac{1}{4}+\frac{t}{2}+\frac{e^{-2t}}{4}\]Yeah, right now I'm just doing the first one but we can verbatim use the same inverse laplace for the second one with the unit step, so overall we should get

Answer 24

yes but u can add u(t) with each term

Answer 25

Looks good to me
\[f(t) = -\frac{1}{4} +\frac{t}{2}+\frac{e^{-2t}}{4}\]
\[y(t) = f(t) - \mathcal{U}(t-1)f(t-1)\]

Answer 26

Oh my GOOOOODDD I just lost all that LaTeX written up due to a browser issue, one sec

Answer 27

\[\frac{ e^{-s} }{ s^2(s+2) }=\frac{ -e^{-s} }{ 4 s }+\frac{ e^{-s} }{ 2s^2}+\frac{ e^{-s} }{ 4(s+2) }\]

Answer 28

that ok?

Answer 29

Yuh

Answer 30

yay

Answer 31

u're a pro now u can do it :)

Answer 32

\[y(t)=-\frac{1}{4}+\frac{t}{2}+\frac{1}{4}e^{-2t}+\Bigg[-\frac{1}{4}+\frac{(t-1)}{2}+\frac{1}{4}e^{-(t-1)}\Bigg]\mathcal{U}(t-1)\]

Answer 33

Whoops, minus sign in between the first collection of terms and the second big term with the unit step

Answer 34

yeah looks fine

Answer 35

And the exponent near the end, \[e^{-2(t-1)}\]

Answer 36

Alright, thank you!

Answer 37

yeah !! :)

Answer 38

really enjoy solving problems with you lulz :)

Answer 39

Lol, ty! You're very helpful.

Answer 40

well i'm happy to be helpful :)