Initial Value Differential Equation Problem

[Wildcat04]

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.

 

[Defennder]

Is there another expression for y_2 ' which you didn't include?

 

[Wildcat04]

Yes Defender, I have added it to my original post.

I am just not sure what the first step would be at this point.

 

[tiny-tim]

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

y₁(0) = 2
y₂(0) = -1

Hi Wildcat04!

hmm … hint: eigenvectors!

 

[Wildcat04]

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 haven't figured out how to make them look nice so I am using TI89 notation :shy:

y' = [[1 4][1 1]] [y₁ y₂]

det => \lambda = 3, -1

y_homo = c₁ [[2][-1]] e^3t + c₂ [[2][-1]] e^-t

Now for the particular solution

g₁ = t² + 6t
g₂ = -t² + t - 1

y_p of the form:

(at² + bt + c)
(dt² + et + d)

Am I heading the right direction?

 

[tiny-tim]

y_homo = c₁ [[2][-1]] e^3t + c₂ [[2][-1]] e^-t

Are they both [[2][-1]]?

Now for the particular solution

g₁ = t² + 6t
g₂ = -t² + t - 1

y_p 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.

 

[HallsofIvy]

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+ t<sup>2[/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).</sup>

 

[Wildcat04]

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

 

[Wildcat04]

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?

\lambda = 3, -1

y(t)_homogen = c₁[[1][2]]e^3t + c₂[[1][-2]]e^-t

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

\varsigma ⁽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 + (c₁e^3t)/3 + (c₂e^-t)/3

y₂' + y₂ = 1/(5)^.5t² + 4/(5)^.5t

y₂(t) = 1/(5)^.5t² + 4/(5)^.5t - c₁e^3t -c₂e^-t


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

Thanks!

 

[HallsofIvy]

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