Question

use Laplace transform to solve the following system:

Answer 1

\[x''=x+3y\]\[y''=4x-4e ^{t}\]initial condition x(0)=2, x'(0)=3, y(0)=1, y'(0)=2

Answer 2

which grade question is it?

Answer 3

hmm from where are you doin, cause we are on this topic now, and im also in 2nd sem, i

Answer 4

lol. i dont think it matters. anyway, i'll try to approach this 1st, but i dont think my notes are sufficient. still searching from the net.

Answer 5

like you must know that how to open for dy/dx, in terms of s and phie

Answer 6

I think i can do this one though

Answer 7

lol, i feel shooo sad, he used 3hours to finish this topic, and im blurring my eyes out...

Answer 8

Lol that sucks being you ehh

Answer 9

so my 1st step, will look like this?\[L[x'']=L[x]+3L[y]\]\[s ^{2}X-2-3=X+3Y\]\[X=\frac{3Y+5}{s ^{2}+1}\]is it correct?

Answer 10

yup alrite

Answer 11

wait its wrong

Answer 12

(X+3Y)/s^2
=first step

Answer 13

sorry you were right , except in the denominator it will be s^2 only

Answer 14

ok i ll do it after lunch

Answer 15

hhehe pagal hai dono

Answer 16

oki will try

Answer 17

I have yet to get a grip on how to solve a system of diffyQs

Laplaces I can manage tho if htats all we really need to do, at least for a first step

Answer 18

: e^-st x'(t) dt = e^-st x(t) + s : e^-st x(t) dt
= - x(0) + s X(s)

: e^-st x''(t) dt = e^-st x'(t) + s : e^-st x'(t) dt
= - x'(0) -sx(0) +s^2 X(s)


the y'' is the same except we replace xs for ys

Answer 19

Lx(t)+3Ly(t) = X(s) + 3Y(s)

4Lx(t)−4Le^t = 4X(s) - 4/(s-1)

Answer 20

s^2 X -2s-3 = X + 3Y

s^2 X -2s-3 = X + 3Y

s^2 X - X = 3Y+2s+3

X(s^2-1)= 3Y+2s+3

X= (3Y+2s+3)/(s^2-1)

Answer 21

same rundown for the Y part i believe, just work the algebra

Answer 22

i got the same X, so i substitute it into the 2nd equation? \[s ^{2}Y-s-2=4X-\frac{4}{s-1}\]

Answer 23

answers obtained \[x=e ^{t}+e ^{2t}\]\[y=e ^{2t}\]