Laplace with boundary conditions

[willmoore21]

Here's the question:

Use laplace transforms to find X(t), Y(t) and Z(t) given that:

X'+Y'=Y+Z
Y'+Z'=X+Z
X'+Z'=X+Y

subject to the boundary conditions X(0)=2, Y(0)=-3,Z(0)=1.

Now I have learnt the basics of laplace transforms, but have not seen a question in this form before. How do I start the question, could someone for instance show me how to get X(t) and I'll try the rest knowing how to do it? I have other questions I need to do like this, but this looks like the easiest one.

Thanks

 

[Deleted member 4993]

Here's the question:

Use laplace transforms to find X(t), Y(t) and Z(t) given that:

X'+Y'=Y+Z
Y'+Z'=X+Z
X'+Z'=X+Y

subject to the boundary conditions X(0)=2, Y(0)=-3,Z(0)=1.

Now I have learnt the basics of laplace transforms, but have not seen a question in this form before. How do I start the question, could someone for instance show me how to get X(t) and I'll try the rest knowing how to do it? I have other questions I need to do like this, but this looks like the easiest one.

Thanks

One way:

X' + Y' + Z' = X + Y + Z

THEN

Z' = X

Z = Y' → X = Y"

and

Y = X' → X = X"'

NoW solve the ODE X - X"' = 0 by LT using BCs X(0) = 2, X'(0) = Y(0) = -3 ....etc..

 

[willmoore21]

I don't really understand this way.

I have found out that I can take the laplace tranform of each line and find X(t) in terms of x(s) the transforms.

So (take L() to mean transform of)

This is the first line's transforms.
L(X')=sx(s)-X(0)
L(Y')=sy(s)-Y(0)
L(Y)=y(s)
L(Z)=z(s)

so I now have

sx(s)-X(0)+sy(s)-Y(0)=y(s)+z(s).

Where do I go from here to solve?

 

[willmoore21]

One way:

X' + Y' + Z' = X + Y + Z

THEN

Z' = X

Z = Y' → X = Y"

and

Y = X' → X = X"'

NoW solve the ODE X - X"' = 0 by LT using BCs X(0) = 2, X'(0) = Y(0) = -3 ....etc..

If I try this way, I get,

L(X)=x(s)
L(X''')=s³x(s)-s²X(0)-sX'(0)-X''(0)

So (colour to illustrate different transforms)

x(s)+s³x(s)-s²X(0)-sX'(0)-X''(0) =0
So I know that X(0)=2, and X'(0)=-3, does X''=1?

this would leave me

x(s)+s³x(s)-2s²+3s-1=0

I have a feeling this is incorrect as I can't just simply sub in. If I just get a little more help on how to get this one, I may be able to do the rest.

Thanks

 

[Deleted member 4993]

If I try this way, I get,

L(X)=x(s)
L(X''')=s³x(s)-s²X(0)-sX'(0)-X''(0)

So (colour to illustrate different transforms)

x(s)+s³x(s)-s²X(0)-sX'(0)-X''(0) =0
So I know that X(0)=2, and X'(0)=-3, does X''=1?

this would leave me

x(s)+s³x(s)-2s²+3s-1=0

I have a feeling this is incorrect as I can't just simply sub in. If I just get a little more help on how to get this one, I may be able to do the rest.

Thanks

x(s)+s³x(s)-2s²+3s-1=0

x(s) = (2s² - 3s + 1)/(1+s³) = A/(1+s) + (Bs + C)/(s² - s + 1)

Solve for A, B and C using usual techniques of partial fraction and continue.....