Laplace transformation: system of DE

[skrat]

Homework Statement

Let $y_1^{'}+y_1=y_2$, $y_2^{'}+5y_2=y_3$, $y_3^{'}+y_3=f$ and $y_1(0)=y_2(0)=y_3(0)=0$. Find $Y_1(s)$ in terms of $F(s)$.

Homework Equations

The Attempt at a Solution

I am completely lost here. I tried to rewrite the system so that I would somehow get rid of $y_2$ and $y_3$ but that didn't really work out well. I don't know what to do.

 

[Ray Vickson]

Homework Statement

Let $y_1^{'}+y_1=y_2$, $y_2^{'}+5y_2=y_3$, $y_3^{'}+y_3=f$ and $y_1(0)=y_2(0)=y_3(0)=0$. Find $Y_1(s)$ in terms of $F(s)$.

Homework Equations

The Attempt at a Solution

I am completely lost here. I tried to rewrite the system so that I would somehow get rid of $y_2$ and $y_3$ but that didn't really work out well. I don't know what to do.

Let $Y_i(s)$ be the Laplace transform of $y_i(t),$ for $i = 1,2,3$. Develop and solve the equations for $Y_1, Y_2, Y_3$. (That is one of the standard methods for dealing with constant-coefficient linear DE systems.)

 

[BiGyElLoWhAt]

Homework Statement

Let $y_1^{'}+y_1=y_2$, $y_2^{'}+5y_2=y_3$, $y_3^{'}+y_3=f$ and $y_1(0)=y_2(0)=y_3(0)=0$. Find $Y_1(s)$ in terms of $F(s)$.

I am completely lost here. I tried to rewrite the system so that I would somehow get rid of $y_2$ and $y_3$ but that didn't really work out well. I don't know what to do.

Looks to me like you have to follow down the line:
if:
$y_{1}' +y_{1} = y_{2}$
and $y_{2}' + 5y_{2} = y_{3}$
then via substitution:

$(y_{1}' + y_{1})' + 5(y_{1}' + y_{1}) = y_{1}'' + (y_{1}' + 5y_{1}') + y_{1} = y_{3}$
follow this logic and I believe you will get your answer.

 

[skrat]

Looks to me like you have to follow down the line:
if:
$y_{1}' +y_{1} = y_{2}$
and $y_{2}' + 5y_{2} = y_{3}$
then via substitution:

$(y_{1}' + y_{1})' + 5(y_{1}' + y_{1}) = y_{1}'' + (y_{1}' + 5y_{1}') + y_{1} = y_{3}$
follow this logic and I believe you will get your answer.

Doing so brings me to:

$y_1^{'''}+7y_1^{''}+11y_1^{'}+5y_1=f$

Now I am guessing that all that still has to be done is:
$Y_1^{'''}(s)+7Y_1^{''}(s)+11Y_1^{'}(s)+5Y_1(s)=F(s)$

And that's it?

 

[BiGyElLoWhAt]

Well, that's what I was thinking, but I didn't really pay much attention to your notation at the beginning, is
$y_{1}$ really $y_{1}(t)$ ?
If that's the case then I guess they want (as previously mentioned) the Laplace transform of $y_{1}$ in terms of the Laplace transform of F. (judging by F(s)).
If by F you mean f and Y you mean y, then yes I would say that that answer would suffice.

 

[BiGyElLoWhAt]

I'm only just learning about Laplace Transforms currently myself. Sorry if these posts weren't much help

 

[skrat]

Hmmm,

BiGyElLoWhAt, your idea works but not so fast. :D

The problem is that $f^{(n)}(t)=s^nF(s)-\sum_{k=1}^{n}s^{k-1}f^{(n-k)}(0)$ but the problem says nothing about the value of $y^{'}$...

So like Ray said, I firstly have write each equation with laplace transformation and than apply your idea which brings me to:

$Y_1(s)[s^3+7s^2+11s+5]=F(s)$

Is this the final result?

 

[LCKurtz]

Hmmm,

BiGyElLoWhAt, your idea works but not so fast. :D

The problem is that $f^{(n)}(t)=s^nF(s)-\sum_{k=1}^{n}s^{k-1}f^{(n-k)}(0)$ but the problem says nothing about the value of $y^{'}$...

So like Ray said, I firstly have write each equation with laplace transformation and than apply your idea which brings me to:

$Y_1(s)[s^3+7s^2+11s+5]=F(s)$

Is this the final result?

That is almost done (and correct). You haven't finished solving for $Y_1$ in terms of $F$ until you divide both sides by that polynomial in $s$. If your ultimate goal is to finish solving by inverting the $Y_i$, you would want to leave the polynomial in $s$ in factored form. Also don't you need to find the other $Y_i\text{'s}$?