Laplace Initial Value Problem with X and Y

[soniccowflash]

Hello. I have gotten as far as to use the Laplace equation with these formulas, but I am having difficulty getting y and x to relate to each other. If requested, I can post my work, but I am sure it is fraught with mistakes. Help is very much appreciated!

x' + 2y' - x - 2y = e^t
x' - y' + x + y = 0

x(0) = 3, y(0) = 0

 

[soniccowflash]

Somebody please help!

 

[JJacquelin]

Hi !

An quick help to begin :

 

[HallsofIvy]

You are doing just about everything wrong! In the first place, this looks very much like homework so I am going to move it to that section. In the second place, yes, you should post your work.

Now, have you considered using basic "systems of algebraic equations" methods to get two equations, one with x' only, the other with y' only?

For example, if you subtract the second equation from the first, you eliminate x' leaving 3y'- 2x- 3y= e^t or y'= (2/3)x+ y+ e^t. If instead you add twice the second equation to the first, you eliminate y' leaving 3x'+ x= e^t.

I had intended to suggest that you use "systems of d.e." methods, perhaps matrix methods, at this point but now I see that the second equation, 3x'+ x= e^t, or x'= (-1/2)x+ e^t, is an equation in x only. Solve that equation for x, then put it back into the first equation to solve for y.

 

[JJacquelin]

You are doing just about everything wrong! In the first place, this looks very much like homework so I am going to move it to that section. In the second place, yes, you should post your work.

Now, have you considered using basic "systems of algebraic equations" methods to get two equations, one with x' only, the other with y' only?

For example, if you subtract the second equation from the first, you eliminate x' leaving 3y'- 2x- 3y= e^t or y'= (2/3)x+ y+ e^t. If instead you add twice the second equation to the first, you eliminate y' leaving 3x'+ x= e^t.

I had intended to suggest that you use "systems of d.e." methods, perhaps matrix methods, at this point but now I see that the second equation, 3x'+ x= e^t, or x'= (-1/2)x+ e^t, is an equation in x only. Solve that equation for x, then put it back into the first equation to solve for y.

Of course, I know all that.
But the question was to solve the problem WITH THE LAPLACE MEHOD, not with the method that you propose.
I suppose the aim is to learn how to use the Laplace transforms, not to use the usual sustitution method or the matricial method, even if they are simpler.
So, I only show how to start and what it is expected to find at the end, for verification.

 

[soniccowflash]

Yes, it is specifically asking to use the Laplace Transform, not algebraic systems. By using the Laplace Transform you do get an algebraic system though. At first I did make the mistake of looking at it in terms of algebra, but the answer looked very wrong. Not that bad of an idea though..just not Laplace.

And JJacquelin, did you use partial fractions? Thank you for the assistance!

How I started with Laplace:

sX(s) - x(0) + 2 sY(s) - y(0) - x(s) - 2 Y(s) = e^t

sX(s) - 3 + 2sY(s) - 0 - x(s)- 2Y(s) = (s-1)^-1

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

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

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

Right now I'm working on incorporating ^this equation into the second.

 

[soniccowflash]

Ok so I tried working with equations 1 and 2, here's what I got:

sx(s) - x(0) - SY(s) - Y(0) + x(s) - Y(s) = 0

sx(s) -3 - SY(s) - 0 + x(s) - Y(s) = 0

x(s)(s+1) - Y(s)(s+1) = 3

x(s) = (3/s+1) + Y(s)

This is where I put x(s) = (s-1)^-2 + (3/s-1) - 2Y(s) in.

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

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

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

Now..I'm not sure what to do with this^ I guess I should find a common denominator then use partial fractions?

 

[JJacquelin]

I think that you made a mistake somewhere. You are on the good track.
Compare your two equations to mine. The notations are :
F(s)=X(s) and G(s)=Y(s)

 

[soniccowflash]

How did you know there was a mistake? And yup, our equations are the same.

 

[JJacquelin]

I don't know where is the mistake, but I got not the same Y(s) :
X(s) = (9s-8)/((s-1)(3s+1))
Y(s) = (7s-5)/((s-1)²(3s+1))

 

[soniccowflash]

Okay..yes. The problem is that with my Y(s), there is a (s+1) and (s-1) in the denominator, but you have (s-1)^2. So I'm not sure how to get two (s-1)'s in the bottom of Y(s) because:

sx(s) - x(0) - SY(s) - Y(0) + x(s) - Y(s) = 0

sx(s) -3 - SY(s) - 0 + x(s) - Y(s) = 0

x(s)(s+1) - Y(s)(s+1) = 3

x(s) = (3/s+1) + Y(s)

 

[JJacquelin]

It's only a system of two linear equations :