Question

Use laplace transforms to solve y''(t)+4y(t)=theta(t) with y(0)=1 y'(0)=0 where theta(t)= 4t, when t is in [0,1] and theta(t)=4 when t>1

Answer 1

\[L[y'(t)=sY(s)-Y(0)\\L[y''(t)]=s^2Y(s)-sY(s)-Y'(0)\]

Answer 2

start by taking laplace transform through out and solve Y

Answer 3

yes the above property shows u the laplace of y'(t) andy''(t)

Answer 4

now just put them in the equation

Answer 5

\[y''(t)+4y(t)=4t\\s^2Y(s)-sY(s)-Y'(0)+4Y(s)=\frac{ 4 }{ s^2 }\]
now put the initial conditions

Answer 6

Yeah I got that far but I got stuck with the theta. Does this mean I need to solve the same equation two different ways, for 4t and 4?

Answer 7

to solve will I need to take the inverse once I find the laplace transform?

Answer 8

right hand side is an unknown function right ?

Answer 9

right hand side is a function theta that can be either 4t or 4, depending on what t is

Answer 10

okay i see

Answer 11

that's the confusing part haha

Answer 12

\[\large \begin{align} y''(t)+4y(t)&=\theta (t)\\~\\
s^2Y(s)-sY(s)-Y'(0)+4Y(s)&=\Theta(s)\\~\\
Y(s) &= \dfrac{1}{s^2-s+4}\Theta(s)\\~\\

\end{align}\]

Answer 13

next you can try convolution

Answer 14

woah....how did you get that above...? i thought you would split it into two equations and take separate laplace transforms

Answer 15

you will need to work two separate eqns for sure, above setup just postpones the splitting part till evaluating the integral :)

Answer 16

ooooooooooooh gotcha!

Answer 17

\[\large \mathcal{L^{-1}}\left(\dfrac{1}{s^2-s+4} \right) = \frac{2}{\sqrt{15}}e^{t/2}\sin(t\sqrt{15}/2)\]

Answer 18

so there is only one solution....not two separate solutions? I'm sorry that still just throws me for a loop lol

Answer 19

\[\large \begin{align} &y''(t)+4y(t)=\theta (t)\\~\\
&s^2Y(s)-sY(s)-Y'(0)+4Y(s)=\Theta(s)\\~\\
&Y(s) = \dfrac{1}{s^2-s+4}\Theta(s)\\~\\
&y(t)=\frac{2}{\sqrt{15}}\int\limits_0^t e^{u/2}\sin(u\sqrt{15}/2)~\theta(t-u)~\mathrm{d}u
\end{align}\]

Answer 20

plugin the value of theta function and evaluate two times, somehow i feel not using convlution gives a smooth solution now.. :O

Answer 21

oh Lord lol. So use convolution or no?

Answer 22

upto you

Answer 23

ok lol. Thanks for your help!

Answer 24

can you work above messy integral ?

Answer 25

ummmm.....honestly not really lol. But I have alot of free time tomorrow so I was gonna muck through it till I figured it out

Answer 26

sounds good, let me give it to wolfram quick and see what it spits out

Answer 27

okie doke

Answer 28

wolfram doesn't know how to evaluate... try it without convolution :)
http://www.wolframalpha.com/input/?i=%5Cfrac%7B2%7D%7B%5Csqrt%7B15%7D%7D%5Cint%5Climits_0%5Et+%284%28t-u%29*e%5E%7Bu%2F2%7D*%5Csin%28u%5Csqrt%7B15%7D%2F2%29%29+du

Answer 29

ok I'll try lol. If i run into problems tomorrow I will let you know!