Question

Alright, my last hurdle to tackle (Systems of Linear Differential Equations), first problem in the section. Posted below momentarily.

Answer 1

\[\frac{dx}{dt}=-x+y\]

Answer 2

\[\frac{dy}{dt}=2x\]

Answer 3

@Kainui , any way you could help me on this? This is my first time dealing with a system of linear diff eqns in any sense. Should I integrate with respect to one or the other, or divide and substitute in? I'm not really sure.

Answer 4

ok came back for this last one :)

Answer 5

start with diffrentiating the first equation

Answer 6

\[\frac{ dx }{ dt }=-x+y\\ \frac{ d^2x }{ dt^2 }=-1+\frac{ dy }{ dt }=2x-1\]
getting this?

Answer 7

back, sorry, was dealing with somebody IRL

Answer 8

I took a different approach, I took the laplace of both sides of all of them and then substituted in for one and took the inverse laplace and stuff

Answer 9

But yeah, that makes sense. Goodnight!

Answer 10

Alright, taking the Laplace of both sides:
\[sX(s)-x(0)=-X(s)+Y(s);\]\[sY(s)-y(0)=2X(s)\]

Answer 11

(Got distracted by stuff, attempting to solve now)

Answer 12

From (2), \[Y(s)=\frac{2X(s)+y(0)}{s}\]

Answer 13

Substituting into the first equation, \[sX(s)-x(0)=-X(s)+\frac{2X(s)+y(0)}{s}\]

Answer 14

Isolating the X(s) term to one side, \[sX(s)+X(s)-\frac{2X(s)}{s}=\frac{y(0)}{s}+x(0)\]

Answer 15

Now Isolating X(s) entirely,

Answer 16

\[X(s)\bigg[s+1-\frac{2}{s}\bigg]=\frac{y(0)}{s}+x(0)\]

Answer 17

Forgot the initial conditions, posting below momentarily

Answer 18

x(0)=0, y(0)=1

Answer 19

\[\frac{1}{s\bigg[s+1-\frac{2}{s}\bigg]}\]

Answer 20

Gonna be a mistake somewhere in there, lol, either way, this doesn't look pretty. Does this look right so far/do you have any suggestions about how to proceed or how I should've proceeded?

Answer 21

OH

Answer 22

\[\frac{1}{s^2+s-2}\]

Answer 23

\[\frac{1}{(s+2)(s-1)}\]

Answer 24

Alright, know how to move from here.