Solution of x^2y''+xy'-y=0 and Regular Point of (1-x^2)y''-2xy'+2y=0

In summary, the general solution of x^2y''+xy'-y=0 can be obtained by using a series solution y=\sum_{n=0}^{\infty} a_n x^n, where the coefficients a_n can be found by setting the sum to 0 and solving for each term. To show that x=0 is a regular point of (1-x^2)y''-2xy'+2y=0, the coefficients a1 and a0 can be expressed as convergent power series in (x-x0) for this value of x0. The geometric series \sum ar^n can also be used to simplify the solution.
  • #1
latentcorpse
1,444
0
What is the general solution of

[itex]x^2y''+xy'-y=0[/itex]
i tried a series solution [itex]y=\sum_{n=0}^{\infty} a_n x^n[/itex]
and whitteld it down to

[itex]\sum_{n=0}^{\infty} (n^2-1) a_n x^n=0[/itex] but this isn't getting me anywhere






secondly how do i show that x=0 is a regular point of
[itex](1-x^2)y''-2xy'+2y=0[/itex]

the definition of a regular point is that :
x0 is a regular point of y''+a1(x)y'+a0(x)y=0 if the coefficients a1 and a0 can be expressed as convergent power series in (x-x0) for this value of x0.

so if i rearrange our equation into the desired form i get that

[itex]a_1(x)=\frac{-2x}{1-x^2},a_2(x)=\frac{2}{1-x^2}[/itex]
for x0=0, a1=0 and a2=2.

what do i say to round of my argument. something like:

"clearly 0 and 2 can be expressed as a convergent power series in x" or
"clearly a1 and a0 arent singular for this value of x0"
how do i round this off?
 
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  • #2
hi latentcorpse - does you series solution imply all [tex]a_n[/tex] are zero except for a certain n?
 
  • #3
A series solution is not the best way to do this. Are you required to use series?

If so, look closely at lanedance's suggestion. If not, try y= xm for some number m.

For your second question, remember that a geometric series, [itex]\sum r^n[/itex] has sum 1/(1- r).
 
  • #4
ok if i was just using series solution, i see now that [itex]a_1[/itex] will be non zero - is there any way of finding its value or is the solution just going to be [itex]y=a_1x[/itex]?

i amen't required to use series solutions but it was a past paper exam question and i couldn't think of any other way to do it - why did you decide to use that substitution?



i don't see how the geometric sum fits into this last bit either.
 
  • #5
You said you had [itex]\sum_{n=0}^{\infty} (n^2-1) a_n x^n=0[/itex].

For Every n, either [itex]a_n= 0[/itex] or [itex]n^2- 1= 0[/itex].

The reason I suggested xm is that (xm)'= mxm-1 and (xm)"= m(m-1)xm-1. so that x(xm)'= mxm and x2(xm)"= m(m-1)xm so every term has the same power of x.

Finally, since the sum of the geometric series [itex]\sum ar^n[/itex] is [itex]a/(1- r)[/itex], [itex]2/(1-x^2)[/itex] can be written as [itex]\sum 2(x^2)^n= \sum 2x^{2n}[/itex]. [itex]2x/(1- x^2)[/itex] is just that multiplied by x: [itex]\sum 2x^{2n+1}[/itex].
 

1. What is a solution to the equation x^2y''+xy'-y=0?

A solution to this equation is a function y(x) that, when substituted into the equation, satisfies the equation for all values of x. In other words, the function makes the equation true.

2. How do you find the general solution to x^2y''+xy'-y=0?

To find the general solution, you first need to solve for y'' using the substitution u=y'. Then, you can solve for y' using the substitution v=y. Finally, you can solve for y using the substitution w=ln(x). This will result in a general solution in the form of y(x)=C1x+C2xln(x), where C1 and C2 are constants.

3. What is a regular point in the equation (1-x^2)y''-2xy'+2y=0?

A regular point in this equation is a point where the coefficients of y'' and y' are not equal to zero. This means that the equation is well-behaved and has a unique solution at that point.

4. How do you determine if a point is a regular point in (1-x^2)y''-2xy'+2y=0?

To determine if a point is a regular point, you can plug it into the equation. If the coefficients of y'' and y' are not equal to zero, then the point is a regular point.

5. What is the relationship between regular points and solutions in (1-x^2)y''-2xy'+2y=0?

Regular points play an important role in determining the existence and uniqueness of solutions in this equation. If a point is a regular point, then a unique solution exists at that point. If a point is not a regular point, then the solution may not be unique or may not exist at all.

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