differential equation: reduction of order technique

Thread Starter

PG1995

Joined Apr 15, 2011
816
Hi

Please have a look on the attachment. You can find my queries there. Please help me with them. I would be very much grateful.

Regards
PG
 

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steveb

Joined Jul 3, 2008
2,436
Hi

Please have a look on the attachment. You can find my queries there. Please help me with them. I would be very much grateful.

Regards
PG
Don't worry about Q1. The author is just using redundant language and trying to make an obvious statement. I agree that the wording is strange, but it is nothing critical, so don't worry about it.

For Q2, the author's solution is correct because \(w={{c}\over{x^6}}\). You can verify this yourself by trial.

However, since \(u'=w\), the the solution for u is \(u={{-c_1}\over{5x^5}}+c_2\).

For Q3, I guess that choice for the constants removes the first solution that you already know and makes the second solution obvious. I agree this is strange and seems to give bias to one of the many possible solutions.

The final solution is very general and works for any values of \(c_1\) and \(c_2\).

There is a simple way to get these solutions, by the way.

If you are adept enough to guess the form of the solution might be \(y=cx^n\), you can get there very quickly, as shown below.

\(y'=cnx^{n-1}\)
\(y''=cn(n-1)x^{n-2}\)

Hence. \(xy''=cn(n-1)x^n\)

Thus, the differential equation gives \(cn(n-1)x^n-6cx^n=0\).

Dividing by\( cx^n \) gives \(n(n-1)-6=0\)

Or, \( n^2-n-6=0\)

Simplify to \((n-3)(n+2)=0\), which means n=3 and n=-2 are the suitable values.

Clearly, the individual solutions are \(y=c_3x^3\) and \(y={{c_{-2}}\over{x^2}}\). It's not hard to see that the addition of both solutions is also a solution. Normally this additive property is something we know holds true for linear equations, but it is not generally true for nonlinear equations, so be careful.

This method of solution is a simple version of the very powerful method of finding a series solution of the form

\(y=\sum_{k=-\infty}^\infty c_i x^i=... +c_{-3}x^{-3}+c_{-2}x^{-2}+c_{-1}x+c_{0}+c_{1}x+c_{2}x^{2}+c_{3}x^{3}+ ...\).

In this simple example, the \(c_2\) and \(c_{-3}\) values are independent and any values will work, but there are equations that might force c1 and c2 to have a fixed ratio between them. Also, this case has only 2 power terms, but other equations might require more terms, or even an infinite number of terms. Many of the famous well known equations/solutions were developed by mathematicians applying this method and figuring out the constraints that determine the correct values of the \(c_i\) coefficients. If this seems confusing, don't worry too much about this now, but it's good to have some idea of this as you go forward.
 
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Thread Starter

PG1995

Joined Apr 15, 2011
816
Thank you very much, Steve.

Thanks for showing me the alternative method. I'm much grateful. Sadly, I have to use the method given in the book.

In the attachment, you can see another example where the author simply make a guess at the values for c1 and c2. I don't understand his criterion for choosing these specific values. But see if you can make some sense out of his method. Thank you.

By the way, the author also gives this formula to find the second solution.

Best regards
PG
 

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steveb

Joined Jul 3, 2008
2,436
... You can see another example where the author simply make a guess at the values for c1 and c2. I don't understand his criterion for choosing these specific values. But see if you can make some sense out of his method.

PG
It seems to me that he is not guessing at the values of those constants, but is instead doing something very specific. He chooses one constant to be zero in order to remove the known solution and isolate the new solution. Then he chooses the other constant to make the other solution have a scale factor of one in front.

The former choice is clearly needed but the latter choice is arbitrary.
 
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