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Differential Equation, finding values

  • Thread starter KCEWMG
  • Start date
  • #1
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Question:
Find the values of ω for which y=cosωt satisfies d^2y/dt^2 +81y = 0
where ω1<ω2


I guess I'm just having a lot of problem even figuring out how to do these at all. Since it's a second degree problem, I found the derivative of y=cosωt, which is -sin(ωt)*ω, and I found the derivative of that to be cos(ωt)*ω. I then set ωcos(ωt)=81cos(ωt).
Am I going about this right at all?
 

Answers and Replies

  • #2
I like Serena
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Welcome to PF, KCEWMG! :smile:

You're doing fine, except that you dropped an omega factor from your 1st to your 2nd derivative.
After correcting that, divide both sides of your equation by cos wt.
 
  • #3
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Alright!

So instead of what I had, it is instead -sin(tω)-tωcos(ωt)=81cos(ωt)
Then I get the -tan(ωt)-tω=81
This might be basic algebra, but where do I go from here? I don't think that I can take an ω out because it's the tangent of ω, right?
 
  • #4
I like Serena
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Errr... you have mangled your 2nd derivative now, I'm afraid.
You almost had it right before.

What did you do to get the derivative of -sin(ωt)*ω?
 
  • #5
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Well, I did it again and got a new answer...
After using the product rule, I got ((cos(ωt)ω)ω-sin(ωt)=81cos(ωt)
Which gets me ω^2=81+sin(ωt)
Right?
 
  • #6
I like Serena
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Let's focus on the derivative first.
Since you're differentiating with respect to t, the derivative of ω is not 1 but 0.
 
  • #7
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OH!! So the final answer is positive and negative 9!
I understand. Thank you!!
 
  • #8
I like Serena
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True! :smile:
 
  • #9
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Alright, the next problem is to find the values of K for which y=x^2+k is a solution to the differential equation 2y-xy'=14.
After that, I figured that 2(x^2+k)-x(2x+k)=14.
I then distributed and solved that k = 14/(2-x).
I put K into the equation y=x^2+(14/2-x).
After doing all of this, I got that y=0.
Now what?
 
  • #10
I like Serena
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Let's focus on the derivative of y first.
You seem to have made the same mistake as in the previous problem...
 

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