Question

Solve DE with undetermined coefficients
y''+9y=t^2e^(3t)+6

Answer 1

Characteristic equation solution = c1cos(3t) + c2sin(3t)

Assume 2 solutions, right?
What are their forms?

Answer 2

Y1 = At^2e^(3t)
Y2 = B

correct? no?

Answer 3

Not correct
Y1 may equal to At^2e^(3t) + Bte^(3t) + Cte^(3t) + D

Answer 4

Oh gosh this is too much.

Answer 5

oh no \(Y_1= (At^2 +Bt + C)e^{3t}\)

Answer 6

\(Y_1'= 3e^{3t}(At^2+Bt+C) +e^{3t}(2At+B)\) not bad :)

Answer 7

Now, you do hehehe. what is \(Y_1"\)?

Answer 8

better take the constant too, otherwise we will have constant 6 on right and no constant on left :P
\(Y_1= (At^2 +Bt + C)e^{3t}+ D\)

Answer 9

Haha. @Loser66
Out of Undermined Coefficients, Laplace Transforms or Variations of Parameters, what would be the quickest method to solve this equation?

Answer 10

Though you didn't ask me, but I am very much tempted to answer that! :P

Laplace all the way!

Answer 11

Haha, I'd love your input as well!!

Answer 12

@hartnn :)

Answer 13

What does one do when initial conditions aren't given?

Answer 14

Assume them to be = 0

Answer 15

Whatever, we have to know all. If the question asks you to use Undetermined coefficient, do it. period

Answer 16

Oh?

Answer 17

(the question doesn't do Undetermined Coefficients).
We have 10 minutes to solve questions like this in class. I thought Undermined would be fastest, but I'm severely regretting it!

Answer 18

\(\color{#0cbb34}{\text{Originally Posted by}}\) @Loser66
Whatever, we have to know all. If the question asks you to use Undetermined coefficient, do it. period
\(\color{#0cbb34}{\text{End of Quote}}\)

Answer 19

Hahaha, you all are the best.

Answer 20

hahahah..... I am not a boss I am a LOSER

Answer 21

I'd say, use the method you have practiced the most with!
I always preferred Laplace over other methods, and can do any ODE within few minutes using laplace, just because I have enough practice of it.

Answer 22

10mins? OK!! Enough

Answer 23

Eh, I'm prone to make algebraic errors.

However, partial fractions can result in errors as well.

Answer 24

How about Variation parameter?

Answer 25

in undetermined co-efficient method, if your initial guess is wrong, you've wasted the time :P so you may avoid it.

Answer 26

Is it easier?
I just imagined taking that my particular solution would have the integral of trigonometric functions multiplied by that nonhomogenous term would be terrible!

Answer 27

Ahhh so I have \[Y(s) = \frac{ \frac{ 6 }{ s } +\frac{ 4 }{ (s-3)^3 }}{ s^2 -9}\]

On the right track?

Answer 28

Err rather, s^2+9 on the denominator

Answer 29

1. s^2 + 9
2. 2/(s-3)^3

Answer 30

Are you sure?

Answer 31

yes :3
http://www.wolframalpha.com/input/?i=laplace+transofrm+of+:+t%5E2e%5E(3t)%2B6

Answer 32

2! = 2

Answer 33

Oh! Duh! Sorry!

Answer 34

this partial fraction is terrible :O

Answer 35

Yeah
:(

Answer 36

*Abort mission*
Go back to Undetermined co-eff :P

Answer 37

I am very confused at how anyone sane can solve this under 10 minutes.

Answer 38

Hahaha

Answer 39

\(Y_1= (At^2 +Bt + C)e^{3t}+ D\)
Y', Y''
then compare the co-efficients on both sides

Answer 40

How can one quickly differentiate y1?

Answer 41

1. Practice
2. Copy from someone nearby :P

Answer 42

Hahahahhha

Answer 43

Will do manually

Answer 44

wise choice :)

Answer 45

You can do it in 10 minutes for sure.

Answer 46

especially since direct comparison of co-efficient gives, B= 0, C= 0, 9D = 6, 18A =1
not too bad!

Answer 47

This makes me think,

\(\color{#0cbb34}{\text{Originally Posted by}}\) @sh3lsh
Y1 = At^2e^(3t)
Y2 = B

correct? no?
\(\color{#0cbb34}{\text{End of Quote}}\)

^^ should have worked

Answer 48

Hmmm

Answer 49

Perhaps we're doing undermined Coefficients incorrectly (or rather, doing extraneous steps?)

Answer 50

I thought
Y = At^2 e^(3t) +B
was already tried...it would have taken so much less time

Answer 51

Is this correct:
If the non homogeneous term is simply a power of t, we must try itself and the lower powers say,
= t^3
Must try At^3 +Bt^2 + Ct + D
If it's a power of t multiplied by exp, just try itself
=t^3e^(3t)
Try At^3e^(3t)

(of course, multiply guess by t if it's in the characteristic solution)

Answer 52

Or must we try the lower powers of t^3e^(3t)

Answer 53

Yes, though i would not use the word "must"
we can try whatever the hell we want!

and here comes the ODE champion! :P

Answer 54

hey how did you get B=0; I have gotten B=-1/27
I know your A is right
And I haven't checked the other constants yet

Answer 55

oh?
i relied on wolfram alpha...

Answer 56

I have to head out!

Thanks all!

Answer 57

\[y=e^{3t}(At^2+Bt+C)+D \text{ should work for the particualr solution }\]