Series Solution to ODEs: Solving y''-6xy'+(6x^2-2)y=0 using Power Series

[kape]

y''-6xy'+(6x^2-2)y=0

y_{1} = _____________

I have to solve the above equation using power series.. but I am stuck. What I have so far is:

y=\sum_{m=0}^\infty a_{m}x^{m}

y'=\sum_{m=1}^\infty ma_{m}x^{m-1}

y''=\sum_{m=2}^\infty m(m-1)a_{m}x^{m-2} = \sum_{m=0}^\infty (m+2)(m+1)a_{m+2}x^{m}

therefore

\sum_{m=0}^\infty (m+2)(m+1)a_{m+2}x^{m} - 6\sum_{m=1}^\infty ma_{m}x^{m} + 6\sum_{m=2}^\infty a_{m-2}x^{m} - 2\sum_{m=0}^\infty a_{m}x^{m} = 0

(and this is where I'm really not sure about)

and because \inline x^{0} occurs in the first and last series..

2a_{2} - 2a_{0} = 0

2(a_{2} - a_{0}) = 0

and \inline x^{1} occurs in the first, second and last series so..

6a_{3} - 8a_{1} = 0

but.. how am i supposed to find the a's from here? Am I supposed to do something like the following?

y = 2(a_{2} - a_{0}) + (6a_{3} - 8a_{1}) ...

I've looked at the textbook and sample problems but I just don't understand..

Can anyone help me??

 

[HallsofIvy]

y''-6xy'+(6x^2-2)y=0

y_{1} = _____________

I have to solve the above equation using power series.. but I am stuck. What I have so far is:

y=\sum_{m=0}^\infty a_{m}x^{m}

y'=\sum_{m=1}^\infty ma_{m}x^{m-1}

so, changing index as you correctly do in y" below
y'= \sum_{m=0}^\infy (m+1)a_{m+1}x^m

y''=\sum_{m=2}^\infty m(m-1)a_{m}x^{m-2} = \sum_{m=0}^\infty (m+2)(m+1)a_{m+2}x^{m}

therefore

\sum_{m=0}^\infty (m+2)(m+1)a_{m+2}x^{m} - 6\sum_{m=1}^\infty ma_{m}x^{m} + 6\sum_{m=2}^\infty a_{m-2}x^{m} - 2\sum_{m=0}^\infty a_{m}x^{m} = 0

No,
\sum_{m=0}^\infty (m+2)(m+1)a_{m+2}x^{m} - 6\sum_{m=0}^\infty (m+1_a_{m+1}x^{m} + 6\sum_{m=2}^\infty a_{m-2}x^{m} - 2\sum_{m=0}^\infty a_{m}x^{m} = 0
where I've just made the change in y'

(and this is where I'm really not sure about)

and because \inline x^{0} occurs in the first and last series..

2a_{2} - 2a_{0} = 0

2(a_{2} - a_{0}) = 0

Not quite. Since m= 0 occurs in the second sum as well,
2a_2- 6a_1- 2a_0= 0
or just
a_2= 3a_1+ a_0
so that a_2 is written in terms of a_0 and a_1.

and \inline x^{1} occurs in the first, second and last series so..<br /> <br /> 6a_{3} - 8a_{1} = 0

<br /> Again, not quite. Taking m= 1 gives<br /> 6a_3- 12a_2- 2a_1= 0<br /> which you can also write as <br /> 6a_3- 12(3a_1+a_0)- 2a_1= 0 <br /> and then <br /> a_3= (17/3)a_1+ 2a_0<br /> which doesn&#039;t look so pretty but is again in terms of a_1 and a_0.<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> but.. how am i supposed to find the a&#039;s from here? Am I supposed to do something like the following? <br /> <br /> y = 2(a_{2} - a_{0}) + (6a_{3} - 8a_{1}) ...<br /> <br /> I&#039;ve looked at the textbook and sample problems but I just don&#039;t understand.. <br /> <br /> Can anyone help me?? </div> </div> </blockquote> What you do with the general formula:<br /> (m+2)(m+1)a_{m+2}- 6(m+1)a_{m+1}+ 6a_{m-2}- 2a_m= 0<br /> depends largely on what you <b>can</b> do!<br /> It&#039;s not that difficult to set m= 2, 3, 4, ,5,... and do the calculations:<br /> with m= 2<br /> 12a_4- 18a_3+ 6a_0- 2a_2= 0<br /> so<br /> a_4= (3/2)a_3- (1/2)a_0- (1/6)a_2<br /> Now, substitute for a_3 and a_2 from above to get<br /> a_4 also in terms of a_0 and a_1.<br /> <br /> Doing that you can get formulas for the first 5 or 6 terms depending upon a_0 and a_1. Often, that is <b>all</b> you can do but sometimes you can &quot;guess&quot; the general formula and then prove it with induction.<br /> <br /> Of course, a_0 and a_1 are the &quot;undetermined constants&quot; you expect to get from a second order differential equation. They can be determined from the initial conditions: a_0= y(0) and a_1= y&amp;#039;(0).

 

[kape]

Thank you for you detailed answer!

But I am still a little confused about the part where you change the index for y'.. why did you change the index for y'?

Since it is 6xy' wouldn't I have to do add another x to the the series? So that it becomes:

y&#039; = \sum_{m=0}^\infty(m+1)a_{m+1}x^{m+1}

In which case, the power of the x's don't match because this one is m+1 so wouldn't I have to change the index back to the original?

y&#039; = \sum_{m=1}^{\infty} ma_{m}x^{m}

I'm afraid I don't have a grasp of the fundamentals, and I find the textbook quite hard to understand. During the process of trying to figure out what is going on by looking at the textbook and sample problems, I think I make incorrect assumptions.. All I understand is that you change the series so that the power of x is all the same, and then try to figure out the undetermined constants.

I also don't quite understand this part. The answer is supposed to be given in the form:

y_{1} = a_{0}( \_\_\_ + \_\_\_x^2 + \_\_\_x^4 ) + a_{1}( \_\_\_x + \_\_\_x^3 ) ...

What is \inline y_{1} ? Is that the same as y'? Is this the "expansion" (terminology?) for the series:

y&#039; = \sum_{m=1}^{\infty} ma_{m}x^{m} ?

I am hopelessly lost and I have so much to do! I hope you reply soon!

Thanks again for all the help.


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(Related Topic)

This is related so I thought you could help me out on this too. The textbook says that:

y&#039; = 2xy

gives

1 \cdot a_{1}x^{0} + \sum_{m=2}^{\infty}ma_{m}x^{m-1} = \sum_{m=0}^{\infty}2a_{m}x^{m+1}

I understand the right side (the 2xy side) but the left side I don't quite get. First of all, why is it m=2 and not m=1? The rest of that series is for y' but for y' shouldn't it be m=1 and not 2? m=2 is for y'', is it not? And the most puzzling of all is this:

1 \cdot a_{1}x^{0}

Where did this come from??

(Incidentally, the dot between 1 and a means multiplication, am I correct? I've seen this in other problems and I don't understand why they explicitly leave it in instead of just multiplying it and showing the results. Why is this? Why, for example, in this case did they not simply show it as \inline a_{1}x^{0}? Is there a purpose to this?)

 

[HallsofIvy]

Thank you for you detailed answer!

But I am still a little confused about the part where you change the index for y'.. why did you change the index for y'?

Since it is 6xy' wouldn't I have to do add another x to the the series?

Oh blast! I misread the equation- I didn't see the "x" in "6xy' ". You are right- that x corrects the exponent so you don't need to change the index.

So that it becomes:

y&#039; = \sum_{m=0}^\infty(m+1)a_{m+1}x^{m+1}

In which case, the power of the x's don't match because this one is m+1 so wouldn't I have to change the index back to the original?

y&#039; = \sum_{m=1}^{\infty} ma_{m}x^{m}

I'm afraid I don't have a grasp of the fundamentals, and I find the textbook quite hard to understand. During the process of trying to figure out what is going on by looking at the textbook and sample problems, I think I make incorrect assumptions.. All I understand is that you change the series so that the power of x is all the same, and then try to figure out the undetermined constants.

I also don't quite understand this part. The answer is supposed to be given in the form:

y_{1} = a_{0}( \_\_\_ + \_\_\_x^2 + \_\_\_x^4 ) + a_{1}( \_\_\_x + \_\_\_x^3 ) ...

What is \inline y_{1} ? Is that the same as y'? Is this the "expansion" (terminology?) for the series:

y&#039; = \sum_{m=1}^{\infty} ma_{m}x^{m} ?

I am hopelessly lost and I have so much to do! I hope you reply soon!

Thanks again for all the help.

I'm not sure why they are calling it y₁ but that is just the general solution to the equation. Let's go back to your original (correct) equation:
\sum_{m=0}^\infty (m+2)(m+1)a_{m+2}x^{m} - 6\sum_{m=1}^\infty ma_{m}x^{m} + 6\sum_{m=2}^\infty a_{m-2}x^{m} - 2\sum_{m=0}^\infty a_{m}x^{m} = 0

As you say, when m= 0, the two "surviving" terms give you 2a_2- 2a_0= 2(a_{2} - a_{0}) = 0 so a_2= a_0.

When m= 1, the three terms give you 6a_3- 6a_1- 2a_1= 6a_3- 8a_1= 0 so a_3= (4/3)a_1.

For m>1, we have all four terms:
(m+2)(m+1)a_{m+2} - 6ma_{m} + 6a_{m-2} - 2a_{m} = 0
which you can solve for a_{m+2}:
a_{m+2}= \frac{(6m+2)a_m- 6a_{m-2}}{(m+2)(m+1)}
What I would now do is make a chart of a_m for as many values of m as I can stand!
a_2= a_0, a_3= (4/3)a_1, a_4= \frac{14}{12}a_2- \frac{6}{12}a_0= \frac{2}{3}a_0, a_5= \frac{20}{20}a_3- \frac{6}{20}a_1= \frac{4}{3}a_1- \frac{3}{10}a_1= \frac{31}{30}a_1, etc.
One thing you can clearly see is that, for even n, a_n will be a multiple of a_0 and, for odd n, a_n will be a multiple of a_1 so the sum breaks easily into even and odd parts:
y(x)= a_0(1+ x^2+ \frac{2}{3}x^2+\cdot\cdot\cdot)+ a_1(x+ \frac{4}{3}x^3+ \frac{31}{30}x^5+\cdot\cot\cot

In fact, once you see that the sum splits into even and odd parts like that, you can simplify the calculation by first taking x_0= 1 and x_1= 0 and then taking x_0= 0 and x_1= 1. (Although, calculating through a_8= \frac{157}{180} and a_9= \frac{109}{1134}, I don't see any obvious pattern!)

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(Related Topic)

This is related so I thought you could help me out on this too. The textbook says that:

y&#039; = 2xy

gives

1 \cdot a_{1}x^{0} + \sum_{m=2}^{\infty}ma_{m}x^{m-1} = \sum_{m=0}^{\infty}2a_{m}x^{m+1}

I understand the right side (the 2xy side) but the left side I don't quite get. First of all, why is it m=2 and not m=1? The rest of that series is for y' but for y' shouldn't it be m=1 and not 2? m=2 is for y'', is it not? And the most puzzling of all is this:

1 \cdot a_{1}x^{0}

Where did this come from??

They have just separated
y&#039;= \sum_{m=1}^\infty m a_m x^{m-1}
into
y&#039;= 1\cdot a_1 x^0+ \sum{m=2}^\infty ma_m x^{m-1}
since a_1 x^0 is just the m= 1 term in the first sum.
The reason they did that (I think) was that then you can change the index in the sum to get
a_1+ \sum_{m=1}^\infty(m+1)a_{m+1}x^m
while the right hand side is
xy= \sum_{m=0}^\infty a_mx^{m+1}[/itex]<br /> and changing the index there gives<br /> \sum_{m=1}^\infty a_{m-1}x^m[/itex]&lt;br /&gt; so that both sums start at m=1 and have x&lt;sup&gt;m&lt;/sup&gt;. What has happened is the m=0 term has already been separated for you.&lt;br /&gt; &lt;br /&gt; &lt;blockquote data-attributes=&quot;&quot; data-quote=&quot;&quot; data-source=&quot;&quot; class=&quot;bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch&quot;&gt; &lt;div class=&quot;bbCodeBlock-content&quot;&gt; &lt;div class=&quot;bbCodeBlock-expandContent js-expandContent &quot;&gt; (Incidentally, the dot between 1 and a means multiplication, am I correct? I&amp;#039;ve seen this in other problems and I don&amp;#039;t understand why they explicitly leave it in instead of just multiplying it and showing the results. Why is this? Why, for example, in this case did they not simply show it as \inline a_{1}x^{0}? Is there a purpose to this?) &lt;/div&gt; &lt;/div&gt; &lt;/blockquote&gt; They just wanted to make it clear (which it wasn&amp;#039;t!) that that term was just ma_mx^{m-1} with m= 1- that&amp;#039;s also why they left x&lt;sup&gt;0&lt;/sup&gt; which is, of course, 1 for all x.