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Need help with method for fourth order DE

  • Thread starter wilkie610
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  • #1
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I was given a 2 part example in class to work on for next class.

1) given the equation y'' + 3y' + 2y = 4 ; Initial conditions are y(0) = y'(0) = 0

I had no problem solving the DE, my result is as follows

y=-4e^(-t) = 2e^(-2t) +2

2) given the equation y'''' + 3y''' + 2y'' = 4 ; Initial conditions are y''''(0) = y'''(0) = y''(0) = 0, a hint was given to substitute z = y'' into the equation to transform it into z'' +3z' +2z = 4. which is similar to the first part of the problem.

My question is what is this method of solving called, the substitution part in number 2? where can i read up on this technique? what do it do next with the z=y''


Thanks
 
Last edited:

Answers and Replies

  • #2
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I think this might be called reduction in order. You might try searching for that term.

In your fourth order equation that you reduced to second order, you have a solution for z, namely z(t) = -4e^(-t) + 2e^(-2t) + 2.

Since z = y'', seems like you should be able to integrate twice to get y.
 
  • #3
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I was thinking that, but wasn't sure, i will give it a try, thanks
 

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