Stuck on the reduction of order step for solving this differential equation

In summary, the conversation is about finding the general solution for the given equation, given that y_1(x) = e^x satisfies the associated homogeneous equation. The conversation also discusses the reduction of order method and the issue of a term involving v not dropping out. The conversation ends with a request for clarification on what is being done wrong.
  • #1
s3a
818
8

Homework Statement


Find the general solution for the equation

(x - 1)y'' - xy' + y = sin(x), x > 1

Given that y_1(x) = e^x satisfies the associated homogeneous equation.


Homework Equations


y_2 = v_2(x) * y_1


The Attempt at a Solution


I read http://tutorial.math.lamar.edu/Classes/DE/ReductionofOrder.aspx and attempted to replicate its method several times and I am attaching my latest attempt. The website I linked to says "Note that upon simplifying the only terms remaining are those involving the derivatives of v. The term involving v drops out. If you’ve done all of your work correctly this should always happen." but I have a term involving v that did not drop out. Also, am I supposed to ignore sin(x) or not? Based on the way the question is phrased, I'd now say I should of ignored it (please tell me if I am correct in saying this) but it doesn't matter for what I am questioning.

By the way, this thread is just about the reduction of order part. (The next step is variation of parameters but I haven't gotten there yet.)

Any help would be greatly appreciated!
Thanks in advance!
 

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  • #2
s3a said:

Homework Statement


Find the general solution for the equation

(x - 1)y'' - xy' + y = sin(x), x > 1

Given that y_1(x) = e^x satisfies the associated homogeneous equation.

Homework Equations


y_2 = v_2(x) * y_1

The Attempt at a Solution


I read http://tutorial.math.lamar.edu/Classes/DE/ReductionofOrder.aspx and attempted to replicate its method several times and I am attaching my latest attempt. The website I linked to says "Note that upon simplifying the only terms remaining are those involving the derivatives of v. The term involving v drops out. If you’ve done all of your work correctly this should always happen." but I have a term involving v that did not drop out. Also, am I supposed to ignore sin(x) or not? Based on the way the question is phrased, I'd now say I should of ignored it (please tell me if I am correct in saying this) but it doesn't matter for what I am questioning.

By the way, this thread is just about the reduction of order part. (The next step is variation of parameters but I haven't gotten there yet.)

Any help would be greatly appreciated!
Thanks in advance!
Check your algebra.

There is no term left involving v.
 
  • #3
I found that the ve^x is supposed to be vxe^x such that they do cancel out (thanks) but now I'm stuck again. Could you please tell me what I am doing wrong now?

If I'm right so far, I don't see how what I did yields y = x.
 

Attachments

  • MyWork.jpg
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  • #4
You got v' correctly,

[tex]v'=C_1e^{-x}(x-1)[/tex].

Integrate, add second constant, multiply by ex.

ehild
 
  • #5
Sorry for the late reply but thanks :).
 
  • #6
Better late than never:wink:

ehild
 
  • #7
Lol ya. :D
 

1. What is the reduction of order method for solving differential equations?

The reduction of order method is a mathematical technique used to solve higher-order differential equations by reducing them to a first-order differential equation. This method is based on the assumption that one solution of the original equation is known, and then a second linearly independent solution can be found using a substitution technique.

2. When is the reduction of order method used?

The reduction of order method is typically used when the coefficients of the differential equation are not constant and cannot be solved using other methods such as separation of variables or the method of undetermined coefficients. It is also useful when one solution of the differential equation is known, and the other solution needs to be found.

3. What are the steps involved in the reduction of order method?

The first step is to find a known solution of the differential equation, denoted as y1. Then, a substitution is made to express the unknown solution, y2, in terms of y1. This substitution is typically in the form of y2 = v(x)y1, where v(x) is a function of x. The next step is to differentiate the substitution and substitute it into the original differential equation. This will result in a first-order differential equation that can be solved for v(x). Finally, the solution for v(x) is substituted back into the substitution to obtain the second solution, y2.

4. Are there any limitations to the reduction of order method?

Yes, there are limitations to this method. It can only be used for linear differential equations, and the known solution, y1, must be linearly independent from the unknown solution, y2. Additionally, if the coefficients of the differential equation are not constant, the substitution may become complicated, making it difficult to solve for v(x).

5. Can the reduction of order method be used for all types of differential equations?

No, the reduction of order method can only be used for second-order linear differential equations. It cannot be used for higher-order differential equations or for non-linear differential equations. In these cases, other methods such as power series or numerical methods may be used to solve the equations.

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