# Homework Help: 2nd Order Diff EQ with 2 intial conditions, got complex roots, i f'ed it up!

1. Feb 9, 2006

### mr_coffee

OKay i havn't gotten 1 2nd order Diff EQ right yet! i'm on a role! wee!!
Find y as a function of t if
81y'' + 126y' + 79y = 0,
y(0) = 2, y'(0) = 9 .

Here is my work:
http://img204.imageshack.us/img204/4605/lastscan5ag.jpg [Broken]

I submitted this and it was wrong!
http://cwcsrv11.cwc.psu.edu/webwork2_files/tmp/equations/dd/f306e7661e8e096c8df72710cdf8c21.png [Broken]

Last edited by a moderator: May 2, 2017
2. Feb 9, 2006

### arildno

Why have you kept your minus inside your sine function?
It's an odd function, simplify your expression a bit.

3. Feb 9, 2006

### mr_coffee

Thanks for the responce, but I fixed it and still ddin't like the answer:
http://cwcsrv11.cwc.psu.edu/webwork2_files/tmp/equations/da/3cf156e7f497d032020dd2db1c553b1.png [Broken]

Last edited by a moderator: May 2, 2017
4. Feb 9, 2006

### arildno

Last I looked "minus minus yield plus"...

5. Feb 9, 2006

### mr_coffee

Whoops, it didn't like that either unless i'm messing up on another sign
http://cwcsrv11.cwc.psu.edu/webwork2_files/tmp/equations/06/82cb0df2333f86d332411fe87ea6fb1.png [Broken]

Last edited by a moderator: May 2, 2017
6. Feb 9, 2006

### arildno

Your line 8 is wrong.
After determining your first constant, you should have:
$$y(t)=e^{-\alpha{t}}(2\cos(wt)+c_{2}\sin(wt)),\alpha=\frac{7}{9},$$
and your expression for w.

7. Feb 9, 2006

### mr_coffee

Why do i have to take the derivative of all that? If i already found c1, can't i just eliminate c1, and take the derivative of the expression with c2 in it and solve for c2? It comes out quite messy if i introduce the e^(-7/9)*2cos(wt);

http://img51.imageshack.us/img51/6304/lastscan0ji.jpg [Broken]

I'm looking off this site here:
http://tutorial.math.lamar.edu/AllBrowsers/3401/ComplexRoots.asp [Broken]

and it says:
http://tutorial.math.lamar.edu/AllBrowsers/3401/ComplexRoots_files/eq0019M.gif [Broken]

Now, you’ll note that we didn’t differentiate this right away as we did in the last section. The reason for this is simple. While the differentiation is not terribly difficult, it can get a little messy. So, first looking at the initial conditions we can see from the first one that if we just applied it we would get the following.
http://tutorial.math.lamar.edu/AllBrowsers/3401/ComplexRoots_files/eq0023M.gif [Broken]

In other words, the first term will drop out in order to meet the first condition. The makes the solution, along with its derivative

http://tutorial.math.lamar.edu/AllBrowsers/3401/ComplexRoots_files/eq0024M.gif [Broken]
A much nicer derivative than if we’d done the original solution.

http://tutorial.math.lamar.edu/AllBrowsers/3401/ComplexRoots_files/eq0025M.gif [Broken]

Last edited by a moderator: May 2, 2017
8. Feb 9, 2006

### assyrian_77

That is valid in the tutorial because they got the constant $$c_{1}$$ to be 0 (zero). In your case it is not zero and can hence not be eliminated like you did.

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