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Homogeneous equations

  • Thread starter jimmianlin
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  • #1

Homework Statement


Given a differential equation.
ie. y'' +y' + 1 =0 (THIS IS NOT THE PROBLEM THAT I AM SOLVING)

Homework Equations


No equations


The Attempt at a Solution


The equation above is not what I'm working with, but an example of a differential equation problem that I was working on. Now, I am trying to find a homogeneous solution by guessing a value for y. However, when I make a guess for y and plug it in, the equation gives me 0 = 1. Does this mean the homogenous solution does not exist or did I not make the correct guess?
 

Answers and Replies

  • #2
rock.freak667
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For a differential equation of the form ay''+by'+cy=0 where a,b,c are constants; all solutions are of the form y=erx
 
  • #3
Dick
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For a differential equation of the form ay''+by'+cy=0 where a,b,c are constants; all solutions are of the form y=erx
That's a little over simplistic. E.g. what about y''=0? jimmianlin, what problem are you trying to solve and are you really supposed to solve it by guessing? Guessing y=constant probably won't get you very far.
 
  • #4
Pengwuino
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The equation above is not what I'm working with, but an example of a differential equation problem that I was working on. Now, I am trying to find a homogeneous solution by guessing a value for y. However, when I make a guess for y and plug it in, the equation gives me 0 = 1. Does this mean the homogenous solution does not exist or did I not make the correct guess?
The homogeneous part of your solution solves the homogeneous part of your problem. Your homogeneous solution, y[x], will solve the problem of y''[x] + y'[x] = 0. Remember, the "1" is the inhomogeneous part so it isn't part of what you're trying to solve for your homogeneous solution.
 
  • #5
rock.freak667
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That's a little over simplistic. E.g. what about y''=0?
Doesn't it work for that as well:confused:? The characteristic equation would be r2=0, giving roots 0,0. So y=(Ax+B)e0x→y=Ax+B

same way you'd get from y''=0 to y'=A to y=Ax+B
 
  • #6
Dick
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Doesn't it work for that as well:confused:? The characteristic equation would be r2=0, giving roots 0,0. So y=(Ax+B)e0x→y=Ax+B

same way you'd get from y''=0 to y'=A to y=Ax+B
I'm just saying Ax+B does not have the form e^rx. That's all.
 
Last edited:
  • #7
HallsofIvy
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You got in ahead of me, Dick! But it's a very useful oversimplification! Looking for solutions of the form [itex]e^rx[/itex] leads to the characteristic equation that then leads to other solutions such as polynomials, sine and cosine, and combinations of those with exponentials. rock.freak667 oversimplified his language a little.
 

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