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Differential Equation problem

  • Thread starter cse63146
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  • #1
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Homework Statement



Compute the general solution of y'' + 6y' + 8y = 2t + et

Homework Equations





The Attempt at a Solution



after using determinants, I found the general solution to be y(t) = k1e-2 + k2e-4 + yp

To find yp, I would make it equal to yp = aet find its first and second derivates, and substitute them back into y'' + 6y' + 8y , but what do I do about the 2t?
 

Answers and Replies

  • #2
Cyosis
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You can do some clever guess work. You can make sure that y'' is 0 by taking some linear function. This simplifies your problem to 6y'+8y=2t. Since we took a linear function y' will be a constant so this will reduce your problem to constant+8y=2t. A function of the form b+ct should suffice.
 
  • #3
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would yp = t be a good guess?

since y'p = 1 and y''p = 0
 
  • #4
Cyosis
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But [itex]6+8t \neq 2t[/itex], so no. Try b+ct as suggested in my previous post.
 
  • #5
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how about [tex]\frac{1}{4}t - \frac{3}{16}[/tex]

so 6(1/4) + 8([tex]\frac{1}{4}t - \frac{3}{16}[/tex]) = 3/2 + 2t - 3/2 = 2t

so when considering et, I would let yp =[tex]\frac{1}{4}t - \frac{3}{16} + e^t[/tex] ?
 
  • #6
Cyosis
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The linear term is correct now, however if we plug in e^t we get, e^t+6e^t+8e^t=15e^t!=e^t. So you have to divide by 15.
 
  • #7
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oops

it should be [tex]\frac{1}{4}t - \frac{3}{16} + ae^t[/tex]

and a would be 1/15
 
  • #8
Cyosis
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Correct.
 
  • #9
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Thank you for all your help.
 
  • #10
Cyosis
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You're welcome.
 
  • #11
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4,860

Homework Statement



Compute the general solution of y'' + 6y' + 8y = 2t + et

Homework Equations





The Attempt at a Solution



after using determinants, I found the general solution to be y(t) = k1e-2 + k2e-4 + yp

To find yp, I would make it equal to yp = aet find its first and second derivates, and substitute them back into y'' + 6y' + 8y , but what do I do about the 2t?
Pretty sure you meant to write y(t) = k1e-2t + k2e-4t + yp here.
 
  • #12
452
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Oops (again).

Yeah, I did. Thanks.
 

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