Simple Differential equation with reduction of order

  • Thread starter trap101
  • Start date
  • #1
342
0
use method of reduction of order to find second solution:

t2y''-4ty+6y = 0 , y1(t)= t2


Attempt:

So I've done all the steps, up to the substitution, but I'm having problems with what appears to be a simple linear equation but I can't solve it: Any ways, with w = v' I arrive at:

w't4 = 0

Everything else canceled out. But trying to get an integrating factor with this would lead me to 0.
 

Answers and Replies

  • #2
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
9,559
770
use method of reduction of order to find second solution:

t2y''-4ty+6y = 0 , y1(t)= t2


Attempt:

So I've done all the steps, up to the substitution, but I'm having problems with what appears to be a simple linear equation but I can't solve it: Any ways, with w = v' I arrive at:

w't4 = 0

Everything else canceled out. But trying to get an integrating factor with this would lead me to 0.

Actually, you are OK so far. Divide out the ##t^4## and you have ##w'=0## or ##v''=0##. Integrate it twice and remember when you are done that ##y=t^2v##. You will see what a second independent solution is.
 
Last edited:
  • #3
342
0
Actually, you are OK so far. Divide out the ##t^4## and you have ##w'=0## or ##v''=0##. Integrate it twice and remember when you are done that ##y=t^2v##. You will see what a second independent solution is.

Your right. This working with zero in this scenario is so weird....

I had another quick question with regards to undetermined coefficients:

y''+2y' = 3 + 4sin(2t)

So working out everything I get everything correct, but I'm missing a term in the particular solution. my solution is:

Yp(t) = -1/2cos(2t) - 1/2sin(2t)

I guess the issue is when I'm setting up my guess and setting the expressions equal. so my initial "guess" is:

Yp(t) = A + B cos(2t) + C sin(2t)

obviously after diffeentiating this twice the constant A term disappeared and as well there is no "y" in the differential equation. yet in the solution they got:

3/2 -1/2cos(2t) - 1/2sin(2t) plus the complimentary equation. How do I get the 3/2?
 
  • #4
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
9,559
770
Your right. This working with zero in this scenario is so weird....

I had another quick question with regards to undetermined coefficients:

y''+2y' = 3 + 4sin(2t)

So working out everything I get everything correct, but I'm missing a term in the particular solution. my solution is:

Yp(t) = -1/2cos(2t) - 1/2sin(2t)

I guess the issue is when I'm setting up my guess and setting the expressions equal. so my initial "guess" is:

Yp(t) = A + B cos(2t) + C sin(2t)

obviously after diffeentiating this twice the constant A term disappeared and as well there is no "y" in the differential equation. yet in the solution they got:

3/2 -1/2cos(2t) - 1/2sin(2t) plus the complimentary equation. How do I get the 3/2?

You didn't show your ##y_c## complementary solution. But ##y=A## obviously is a solution of the homogeneous equation, so it can't work for ##y_p##. Try ##y_p=At## plus the sine an cosine terms.
 
  • #5
342
0
You didn't show your ##y_c## complementary solution. But ##y=A## obviously is a solution of the homogeneous equation, so it can't work for ##y_p##. Try ##y_p=At## plus the sine an cosine terms.


Sorry for the delayed response. had class. my complementary solution was:

yc(t) = c1e0t + c2e-2t

So where does the A term appear in the complemetary solution? is it because c1 is only the constant without e0t in it?
 
  • #6
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
9,559
770
Sorry for the delayed response. had class. my complementary solution was:

yc(t) = c1e0t + c2e-2t

So where does the A term appear in the complemetary solution? is it because c1 is only the constant without e0t in it?

If you simplify ##c_1e^{0t}## what do you get?
 
  • #7
342
0
If you simplify ##c_1e^{0t}## what do you get?



Yes. I get the constant alone. Is it because it's the constant or should I look at it as it's because they are the same degree polynomial?
 
  • #8
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
9,559
770
Yes. I get the constant alone. Is it because it's the constant or should I look at it as it's because they are the same degree polynomial?

Too many pronoun "its" in that question. I have no idea what you are asking me.
 
  • #9
342
0
Too many pronoun "its" in that question. I have no idea what you are asking me.


Sorry. What I was asking is that since in the complementary solution I have y(t) = c1 + c2e-2 does the constant c1 match with the "A" that I have as my "guess" for the polynomial portion of the particular solution? and if so, then that's my I add a "t" to it? i.e: At + (the rest)?
 
  • #10
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
9,559
770
Sorry. What I was asking is that since in the complementary solution I have y(t) = c1 + c2e-2 does the constant c1 match with the "A" that I have as my "guess" for the polynomial portion of the particular solution? and if so, then that's my I add a "t" to it? i.e: At + (the rest)?

Yes. The fact that a constant ##c_1## is a solution of the homogeneous equation means that a constant, no matter whether you call it ##A## or something else can't be a solution of the NH equation. So no point in trying ##y_p = A##.
 

Related Threads on Simple Differential equation with reduction of order

Replies
1
Views
1K
Replies
0
Views
2K
Replies
7
Views
20K
Replies
2
Views
4K
Replies
1
Views
4K
Replies
10
Views
2K
  • Last Post
Replies
2
Views
589
  • Last Post
Replies
5
Views
2K
Replies
2
Views
1K
Replies
1
Views
54
Top