Initial Value Differential Equation Problem

  • Thread starter Wildcat04
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  • #1
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Homework Statement



y1' = y1 + 4y2 - t2 +6t
y2' = y1 + y2 - t2 + t -1

y1(0) = 2
y2(0) = -1

Could someone give me a nudge as to how to properly complete this problem? We never really went over anything with multiple y values and I am a little confused as to how I should approach this problem.

Thank you in advance.
 
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Answers and Replies

  • #2
Defennder
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Is there another expression for [tex]y_2 ' [/tex] which you didn't include?
 
  • #3
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Yes Defender, I have added it to my original post.

I am just not sure what the first step would be at this point.
 
  • #4
tiny-tim
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y1' = y1 + 4y2 - t2 +6t
y2' = y1 + y2 - t2 + t -1

y1(0) = 2
y2(0) = -1

Hi Wildcat04! :smile:

hmm :rolleyes: … hint: eigenvectors! :wink:
 
  • #5
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Ok, I should have thought of that...I think that I have found the homogen. part of the solution but I am struggling with the particular solution.

Sorry about the matrices, I havent figured out how to make them look nice so I am using TI89 notation :shy:

y' = [[1 4][1 1]] [y1 y2]

det => [tex]\lambda = 3, -1[/tex]

yhomo = c1 [[2][-1]] e3t + c2 [[2][-1]] e-t

Now for the particular solution

g1 = t2 + 6t
g2 = -t2 + t - 1

yp of the form:

(at2 + bt + c)
(dt2 + et + d)

Am I heading the right direction?



 
  • #6
tiny-tim
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yhomo = c1 [[2][-1]] e3t + c2 [[2][-1]] e-t

Are they both [[2][-1]]?
Now for the particular solution

g1 = t2 + 6t
g2 = -t2 + t - 1

yp of the form:

(at2 + bt + c)
(dt2 + et + d)

Am I heading the right direction?

hmm … i'm a bit lost here …

i expect there is a matrix way of doing it …

but i was just going to separate into the two eigenvectors, and solve for each separately. :redface:
 
  • #7
HallsofIvy
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Or equivalently, a little simpler in concept but harder calculations:

Differentiate y1'= y1 + 4y2 - t2 +6t with respect to t to get y1"= y1'+ 4y2'- t2+ 6t. Since y2'= y1 + y2 - t2 + t -1, that is y1"= y1'+ 4y1- 4y2- 4t2+ 4t- 4- t2+ 6t= y1'+ 4y1- 4y2- 5t2+ 10t- 4.

But from the first equation, 4y2= y1'- y1+ t2- 6t so
y1"= y1'+ (y1'- y1+ t2[/sup- 6t)- 5t2+ 10t- 4 or

y1"= 2y1'- y1- 4t2+ 4t+ 4 or

y1"- 2y1'+ y1= -4t2+ 4t+ 4
with y1(0)= 2 and, since 4y2(0)= -4= y1'(0)- y1(0)+ 02- 6(0)= y1'(0)-2,
y'(0)= -2.

Solve that equation for y1(x) and then use 4y2= y1'- y1+ t2- 6t to solve for y2(x).
 
  • #8
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Going off of Halls of Ivy's equations,

Please let me know if these are correct assumptions

y1'(0) = -2 => y1'(t) = t - 2
y1(0) = 2 => y1(t) = .5t2 - 2t +2

4y2 = y1' + y1 + t2 - 6t
4y2 = t - 2 - .5t2 + 2t - 2 + t2 - 6t
4y2 = -.5t2 - 3t -4

=> y2(t) = -.125t2 - .75t -1
 
  • #9
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Tiny Tim,

For my own sake I also wanted to try and do it using eigenvectors and I think that I have reached a solution (or at least I think that I am close). If you get a chance could you please take a look and let me know if it is correct?

[tex]\lambda = 3, -1[/tex]

y(t)homogen = c1[[1][2]]e3t + c2[[1][-2]]e-t

[tex]\varsigma[/tex] (1) = 1/(12 + 22).5 [[1][2]] = 1/(5).5 [[1][2]]

[tex]\varsigma[/tex] (2) = 1/(5).5 [[1][-2]]

T=(1/(5).5) [[1,1][2,-2]]

T-1 = [[1.2][1,-2]]

=> =>

y1' - 3y1 = -3/(5).5t2 + 8/(5).5t

y1(t) = 1/(5).5t2 + 2.66/(5).5t + (c1e3t)/3 + (c2e-t)/3

y2' + y2 = 1/(5).5t2 + 4/(5).5t

y2(t) = 1/(5).5t2 + 4/(5).5t - c1e3t -c2e-t


Am I completely off base or is this a reasonable solution?

Thanks!
 
  • #10
HallsofIvy
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Going off of Halls of Ivy's equations,

Please let me know if these are correct assumptions

y1'(0) = -2 => y1'(t) = t - 2
Are you assuming that y1"(t)= 1 for all t? Why?
As I said before, y1"- 2y1'+ y1= -4t2+ 4t+ 4

You said before that the problem was "We never really went over anything with multiple y values" so I assume you can solve a d.e. with a single y1.

y1(0) = 2 => y1(t) = .5t2 - 2t +2

4y2 = y1' + y1 + t2 - 6t
4y2 = t - 2 - .5t2 + 2t - 2 + t2 - 6t
4y2 = -.5t2 - 3t -4

=> y2(t) = -.125t2 - .75t -1
 

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