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Solve differential equation using power series

  • #1
Find y(x) as a power series satisfying:
y'-2xy=0 , y(0) = 1

attempt:

y=[itex]\sum[/itex][itex]^{∞}_{n=0}[/itex] an(x-x0)n

y'=[itex]\sum[/itex][itex]^{∞}_{n=0}[/itex] (n+1)an+1(x-x0)n

y' - 2xy = y' - 2y(x-x0) - 2x0y

substituting the power series into above formula and simplifying eventually gives:

a1-2x0a0 + [itex]\sum[/itex][itex]^{∞}_{n=1}[/itex](x-x0)n[(n+1)an+1 - 2an-1 - 2x0an] = 0

so when n=0 and n=1 we get:
a1=2x0a0

when n>1,
an = [itex]\frac{2}{n}[/itex](x0an-1 + an-2)

but what does my professor mean by y(0) = 1? I think he was trying to explain that this means that x0 = 0 ?



Also, another problem is that y' - 2xy = x3ex2, y(0)=1
In this case, any hint as to what should I do with the e^x^2????
 
Last edited by a moderator:

Answers and Replies

  • #2
33,262
4,963
Find y(x) as a power series satisfying:
y'-2xy=0 , y(0) = 1

attempt:

y=[itex]\sum[/itex][itex]^{∞}_{n=0}[/itex] an(x-x0)n

y'=[itex]\sum[/itex][itex]^{∞}_{n=0}[/itex] (n+1)an+1(x-x0)n

y' - 2xy = y' - 2y(x-x0) - 2x0y

substituting the power series into above formula and simplifying eventually gives:

a1-2x0a0 + [itex]\sum[/itex][itex]^{∞}_{n=1}[/itex](x-x0)n[(n+1)an+1 - 2an-1 - 2x0an] = 0

so when n=0 and n=1 we get:
a1=2x0a0

when n>1,
an = [itex]\frac{2}{n}[/itex](x0an-1 + an-2)

but what does my professor mean by y(0) = 1? I think he was trying to explain that this means that x0 = 0 ?
This means that y = 1 when x = 0.
Also, another problem is that y' - 2xy = x3ex2, y(0)=1
In this case, any hint as to what should I do with the e^x^2????
 
  • #3
This means that y = 1 when x = 0.
yes I know that but what I meant was, how can I use that to solve the problem? If I already have the recursive formula giving me the coefficients in terms of an, of what use is that y(0)=1? On my hw it says y(0)=1 but when my professor was talking about it, he wrote y(x0)=1, but they are not the same thing?? Is there some relation between them I am not seeing?

I mean if x=0, then
y(0)=[itex]\sum^{∞}_{n=1}[/itex](0-x0)n=1
so:
1-x0+x02-x03+... = 1
so this implies x0=0 ? but then all of my coefficients for odd n's become 0..idk if thats supposed to happen...
 
Last edited:
  • #4
1,796
53
Also, another problem is that y' - 2xy = x3ex2, y(0)=1
In this case, any hint as to what should I do with the e^x^2????
Nothing wrong with the odd ones being zero if they have to. And as far as that [itex]e^{x^2}[/itex], just expand it out in it's power series form and equate coefficients on both sides just like you did above except in this case, all the coefficients on the right side are no longer zero.
 
  • #5
Nothing wrong with the odd ones being zero if they have to. And as far as that [itex]e^{x^2}[/itex], just expand it out in it's power series form and equate coefficients on both sides just like you did above except in this case, all the coefficients on the right side are no longer zero.
So but then y(0)=1 means x0=0 and that a0 =1, right?
 
  • #6
33,262
4,963
Yes, x0 = 0, so your series should be expanded in powers of x, meaning that your series solutions is y = a0 + a1x + a2x2 + .... The initial condition y(0) = 1 determines a0.
 
  • #7
1,796
53
So but then y(0)=1 means x0=0 and that a0 =1, right?
That is correct. Also, compute four or so of the coefficients and then compare them to that series you wrote for the right side of the other one.
 

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