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 Post subject: y""-y=u1(t)-u2(t)
PostPosted: Thu, 3 Nov 2011 21:29:14 UTC 
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Posts: 261
Here is what I am given:

\y^{(4)}-y=u1(t)-u2(t); y(0)=0, y'(0)=0, y"(0)=0, y"'(0)=0

So here is what I tried:

L{\y^{(4)}}-L{y}=L{u1(t)}-L{u2(t)}

sL{y"'}-y"'-L{y}=L{u1(t)}-L{u2(t)}

s(sL{y"})-y"-L{y}=L{u1(t)}-L{u2(t)}

\s^{2}(sL{y'})-y'-L{y}=L{u1(t)}-L{u2(t)}

\s^{3}(sL{y})-y-L{y}=L{u1(t)}-L{u2(t)}

\s^{4}L{y}-L{y}=L{u1(t)}-L{u2(t)}

(\s^{4}-1)L(y)=L{u1(t)}-L{u2(t)}

Now my problem is that the problem didn't tell me a thing about u1 or u2 so I have no idea what to do with them.


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 Post subject: Re: y""-y=u1(t)-u2(t)
PostPosted: Thu, 3 Nov 2011 21:33:14 UTC 
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Joined: Fri, 30 Oct 2009 16:33:10 UTC
Posts: 261
Oh boy, the first post doesn't look right at all. Hopefully this is a little easier to read:

y^(4)-y=u1(t)-u2(t); y(0)=0, y'(0)=0, y"(0)=0, y"'(0)=0

So here is what I tried:

L{y^(4)}-L{y}=L{u1(t)}-L{u2(t)}

sL{y"'}-y"'-L{y}=L{u1(t)}-L{u2(t)}

s(sL{y"})-y"-L{y}=L{u1(t)}-L{u2(t)}

s^{2}(sL{y'})-y'-L{y}=L{u1(t)}-L{u2(t)}

s^{3}(sL{y})-y-L{y}=L{u1(t)}-L{u2(t)}

s^{4}L{y}-L{y}=L{u1(t)}-L{u2(t)}

(s^{4}-1)L(y)=L{u1(t)}-L{u2(t)}

Now my problem is that the problem didn't tell me a thing about u1 or u2 so I have no idea what to do with them.


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 Post subject: Re: y""-y=u1(t)-u2(t)
PostPosted: Fri, 4 Nov 2011 05:59:35 UTC 
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Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 13529
Location: Austin, TX
goteamusa wrote:
Oh boy, the first post doesn't look right at all. Hopefully this is a little easier to read:

y^(4)-y=u1(t)-u2(t); y(0)=0, y'(0)=0, y"(0)=0, y"'(0)=0

So here is what I tried:

L{y^(4)}-L{y}=L{u1(t)}-L{u2(t)}

sL{y"'}-y"'-L{y}=L{u1(t)}-L{u2(t)}

s(sL{y"})-y"-L{y}=L{u1(t)}-L{u2(t)}

s^{2}(sL{y'})-y'-L{y}=L{u1(t)}-L{u2(t)}

s^{3}(sL{y})-y-L{y}=L{u1(t)}-L{u2(t)}

s^{4}L{y}-L{y}=L{u1(t)}-L{u2(t)}

(s^{4}-1)L(y)=L{u1(t)}-L{u2(t)}

Now my problem is that the problem didn't tell me a thing about u1 or u2 so I have no idea what to do with them.


You cannot do too much about them, you just gotta write L(u_1} et cetera.

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