Simple Differential equation with reduction of order

In summary: Yes. The fact that a constant ##c_1## is...added to the initial guess for the polynomial gives the final answer for that term in the complementary solution.
  • #1
trap101
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.
 
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  • #2
trap101 said:
use method of reduction of order to find second solution:

t2y''-4ty+6y = 0 , y1(t)= t2Attempt:

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
LCKurtz said:
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
trap101 said:
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
LCKurtz said:
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
trap101 said:
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
LCKurtz said:
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
trap101 said:
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
LCKurtz said:
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
trap101 said:
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##.
 

1. What is a simple differential equation with reduction of order?

A simple differential equation with reduction of order is a type of differential equation that can be solved by reducing the order of the equation. This involves finding a new variable to substitute into the original equation, which simplifies it and makes it easier to solve.

2. How do you solve a simple differential equation with reduction of order?

To solve a simple differential equation with reduction of order, you first need to find a new variable to substitute into the equation. This variable should be related to the original variable by a simple algebraic equation. Once you have the new equation, you can solve it using standard techniques, such as separation of variables or integrating factors.

3. What are the benefits of using reduction of order to solve a differential equation?

The reduction of order technique can simplify a complex differential equation, making it easier to solve. It also allows for the use of standard integration techniques, which can be more straightforward and efficient compared to other methods of solving differential equations.

4. Can any differential equation be solved using reduction of order?

No, not all differential equations can be solved using reduction of order. This method is only applicable to a specific type of differential equation, known as a "homogeneous" equation. Other types of equations may require different techniques, such as separation of variables or the use of special functions.

5. What are some real-world applications of simple differential equations with reduction of order?

Simple differential equations with reduction of order have many applications in various fields of science, including physics, engineering, and biology. They are commonly used to model natural phenomena, such as population growth, heat transfer, and chemical reactions. They are also used in the development of mathematical models for predicting and analyzing complex systems.

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