Question

Use laplace to solve the system to differential equations.

dx/dt + 2y - x = 0

dy/dt -2x -y = 0

x(0) =1 and y(0) = 0

I don't need the full answer or anything... Just a guide or quick overview of exactlyy What i'm meant to do would be sufficient...

What i did was, find the laplace of each one. I then put the first one in terms X(t) and and subbed it into the second one, to find the solution of y(t).. But that answer was wrong. thanks guys!

Answer 1

You should start by finding the Laplace transform of each equation, you will get two equations in two unknowns \(X(s)\) and \(Y(s)\). Use one equation to write \(X(s)\) in terms of \(Y(s)\) and substitute in the other one. I'll do the one part of the problem here.
Laplace transform for the two equation gives:
\(sX(s)-1+2Y(s)-X(s)=0 \implies (s-1)X(s)+2Y(s)=1\) \((1)\)
\(sY(s)-2Y(s)-Y(s)=0 \implies (s-1)Y(s)-2X(s)=0\) \((2)\)
From (2) we have \(X(s)=\frac{s-1}{2}Y(s)\). Substitute for X(s) in (1) to get an equation in Y(s); that's, after multiplying by 2, \((s-1)^2Y(s)+4Y(s)=2\), which can be written as \(Y(s)=\frac{2}{(s-1)^2+4}\). Its inverse laplace transform is y(t) is sin2t with s shifted one point to the right. So, \(y(t)=\sin(2t)e^t\).

Answer 2

ggaborne....r u here...

Answer 3

Yes! I understand what AnwarA has said.. And I did what he did at first.. I just did a simple mistake in rearranging... But yay I did it right the first time minus that fact!!

Answer 4

i don't know how anwar got the laplace form of first equation as sX(s)−1+2Y(s)−X(s)=0

Answer 5

the -1 should not be there...

Answer 6

otherwise his approach is right

Answer 7

It's the x(0) = 1

Answer 8

oh.....i missed that part of your question....

Answer 9

that's alright lol. thanks anywya! Im sure ill have more questions soon

Answer 10

How long will you be online for?

Answer 11

sorry for that...if u have any good problems u can post...

Answer 12

2 hrs....

Answer 13

come in the chat box....